## 1. Abridged problem statement

We are given an incomplete chronological log of a club. Each existing record is one of:

- `+ name`: person enters;
- `- name`: person leaves;
- `= k`: exactly `k` people are inside at that moment.

Initially the club is empty. Some records are missing, but the given records must stay in the same relative order.

We must insert as few records as possible so that the final log is physically valid:

- nobody can leave before entering;
- nobody can enter twice without leaving in between;
- each `= k` must match the actual current number of visitors.

Output the minimum possible final log length and one valid completed log.

Constraints: `1 <= N <= 200`, `0 <= k <= 100`.

---

## 2. Detailed editorial

### Key idea

We process the given log from left to right and maintain a plausible current state of the club.

The obvious state would be the exact set of people inside. However, many people may never appear again in the remaining original log. Their identities do not matter anymore — they only contribute to the total count. The solution uses this observation heavily.

We distinguish two kinds of people currently inside:

1. **Named people**: people whose names may appear again later in the original log.
2. **Anonymous people**: people whose names will never appear again, or freshly generated people. We only need to know their names so we can output a matching `- name` later if needed.

We maintain:

- `present[x]`: whether original name `x` is currently inside as a named person;
- `anon`: stack of anonymous names currently inside;
- `cnt`: total number of people inside.

---

### Preprocessing

Compress all names from the input into integer IDs.

For each name, store all positions where it appears in `+` or `-` records. This lets us answer:

> What is the next original occurrence of this name after the current position?

Also store `last_at[x]`, the last position where name `x` appears.

If we process `+ x` and this is the last occurrence of `x` in the input, then after this point the name `x` will never matter again. We can put this person into the anonymous stack instead of keeping them as a named active visitor.

---

### Processing `+ x`

If `x` is already inside, then the original `+ x` is impossible directly: a person cannot enter twice.

So we must insert:

```text
- x
```

before it.

Then we process the original `+ x`.

If this is the last occurrence of `x`, we add `x` as anonymous. Otherwise, we mark `x` as named-present.

This is forced and therefore optimal.

---

### Processing `- x`

If `x` is currently inside, we simply output the original `- x` and mark `x` absent.

If `x` is not inside, then this leave record is impossible unless we insert a matching enter immediately before it:

```text
+ x
- x
```

This is also forced and optimal.

---

### Processing `= k`

This is the only interesting case.

At this point the current number of guests is `cnt`.

If `cnt == k`, we simply output `= k`.

If `cnt < k`, we need to insert exactly `k - cnt` enter records before `= k`.

If `cnt > k`, we need to insert exactly `cnt - k` leave records before `= k`.

The number of inserted records here is forced, but we can choose *which names* to insert/close in a smart way to reduce future insertions.

---

### Case 1: `cnt < k`

We must add visitors.

The best people to add are absent names whose next occurrence is `- x`.

Why?

If the next original record involving `x` is `- x`, then by inserting `+ x` now, that future `- x` becomes valid for free.

So one inserted record now may save one inserted record later.

Among such candidates, choose those whose next `- x` happens earliest.

If we still need more people, generate fresh unused names and add them as anonymous guests.

---

### Case 2: `cnt > k`

We must remove visitors.

There are three categories of current visitors.

#### 1. Named people whose next event is `+ x`

If `x` is currently inside and the next original event is `+ x`, then eventually we must remove `x` before that `+ x` anyway.

So closing such people now is beneficial: the inserted `- x` simultaneously satisfies the current `= k` and prepares for the future `+ x`.

We close these first, ordered by earliest next `+ x`.

#### 2. Anonymous people

They have no future obligations. Closing one costs exactly one inserted record and has no future penalty.

We close these after the previous category.

#### 3. Named people whose next event is `- x`

These are useful to keep, because their future `- x` can happen naturally.

If we close such a person now, then when the future `- x` appears, we will have to insert a new `+ x` before it.

So this is the worst category to close.

If we must close some of them, close the one whose future `- x` is latest, because it is the least urgent/useful one.

---

### Why greedy is optimal

For `+ x` and `- x`, the necessary insertions are forced.

For `= k`, the number of immediate insertions is forced: exactly `|cnt - k|`.

The only question is the identity of inserted/removed people.

The chosen priorities are based on local exchange arguments:

- When increasing count, adding someone whose next event is `- x` can only help, because it makes that future leave valid.
- Earlier future leaves are more valuable because they are used sooner.
- When decreasing count, closing someone whose next event is `+ x` can only help, because we would have needed to close them before that future enter anyway.
- Anonymous guests have no future value.
- People whose next event is `- x` should be preserved as much as possible, because they already have a useful future leave.

Thus the greedy choice always leaves a state at least as good as any alternative for the unprocessed suffix.

Complexity is small: at most `O(N * D log D)`, where `D <= N` is the number of distinct names.

---

## 3. Provided C++ solution with detailed comments

```cpp
#include <bits/stdc++.h>          // Include all standard C++ library headers.

using namespace std;              // Avoid writing std:: everywhere.

// Output operator for pairs. Not essential for the algorithm, but convenient.
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second; // Print pair elements separated by space.
}

// Input operator for pairs. Also not essential here.
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second; // Read pair elements.
}

// Input operator for vectors.
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {               // Iterate over all elements by reference.
        in >> x;                    // Read each element.
    }
    return in;                      // Return stream for chaining.
};

// Output operator for vectors.
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {                // Iterate over all elements.
        out << x << ' ';            // Print each followed by a space.
    }
    return out;                     // Return stream for chaining.
};

int n;                              // Number of original records.
vector<int> rec_type;               // Record type: 0 = '+', 1 = '-', 2 = '='.
vector<int> rec_arg;                // For '+/-': name id. For '=': required count.
vector<string> names;               // names[id] = original string name.

// Reads and compresses the input.
void read() {
    cin >> n;                       // Read number of records.

    rec_type.assign(n, 0);          // Allocate record type array.
    rec_arg.assign(n, 0);           // Allocate record argument array.

    map<string, int> id;            // Maps each seen name to its integer id.

    for(int i = 0; i < n; i++) {    // Read all records.
        char c;                     // First character: '+', '-', or '='.
        cin >> c;

        if(c == '=') {              // Count-check record.
            rec_type[i] = 2;        // Type 2 means '='.
            cin >> rec_arg[i];      // Store required count.
        } else {                    // Enter or leave record.
            rec_type[i] = (c == '+') ? 0 : 1; // Type 0 for '+', type 1 for '-'.

            string nm;              // Name in this record.
            cin >> nm;

            auto it = id.find(nm);  // Check whether the name already has an id.

            if(it == id.end()) {    // New name.
                int k = names.size(); // Next available id.
                id[nm] = k;          // Store id in map.
                names.push_back(nm); // Store string in names array.
                rec_arg[i] = k;      // Store id as argument of this record.
            } else {                 // Existing name.
                rec_arg[i] = it->second; // Reuse old id.
            }
        }
    }
}

void solve() {
    int d = names.size();           // Number of distinct names from the original log.

    vector<vector<int>> occ(d);     // occ[x] = positions where name x appears.

    for(int i = 0; i < n; i++) {    // Scan all original records.
        if(rec_type[i] != 2) {      // Only '+/-' records contain names.
            occ[rec_arg[i]].push_back(i); // Save occurrence position.
        }
    }

    vector<int> last_at(d, -1);     // last_at[x] = last occurrence position of name x.

    for(int x = 0; x < d; x++) {    // For every name.
        if(!occ[x].empty()) {       // If it appears at least once.
            last_at[x] = occ[x].back(); // Last occurrence is the last stored position.
        }
    }

    // Returns the first occurrence position of name x at or after position from.
    auto next_occ = [&](int x, int from) {
        auto& v = occ[x];           // Occurrence list of name x.
        auto it = lower_bound(v.begin(), v.end(), from); // First occurrence >= from.
        return it == v.end() ? INT_MAX : *it; // INT_MAX means no future occurrence.
    };

    set<string> used_names(names.begin(), names.end()); // Original names cannot be reused as fresh names.

    vector<int> digits(1, 0);       // Base-26 counter for generating names: a, b, ..., z, aa, ...

    // Generates a lowercase name not present in the original log.
    auto next_fresh = [&]() {
        while(true) {               // Keep trying until an unused name is found.
            string s;               // Candidate fresh name.

            for(int v: digits) {    // Convert base-26 digits to letters.
                s.push_back((char)('a' + v));
            }

            int p = (int)digits.size() - 1; // Start incrementing from last digit.

            while(p >= 0 && digits[p] == 25) { // Carry over trailing 'z' digits.
                digits[p] = 0;      // Reset this digit to 'a'.
                p--;                // Carry to previous digit.
            }

            if(p < 0) {             // Overflow, e.g. z -> aa.
                digits.assign(digits.size() + 1, 0); // Increase length, all 'a'.
            } else {
                digits[p]++;        // Increment current digit.
            }

            if(!used_names.count(s)) { // If candidate is not an original name.
                return s;           // Use it.
            }
        }
    };

    vector<char> present(d, 0);     // present[x] = whether original name x is inside as a named guest.
    vector<string> anon;            // Stack of anonymous guests' names currently inside.
    int cnt = 0;                    // Total number of people currently inside.
    vector<string> out;             // Final polished log.

    // Add a named guest x to the club.
    auto open_named = [&](int x) {
        out.push_back("+ " + names[x]); // Output enter record.
        present[x] = 1;                 // Mark x as inside.
        cnt++;                          // Increase total count.
    };

    // Remove a named guest x from the club.
    auto close_named = [&](int x) {
        out.push_back("- " + names[x]); // Output leave record.
        present[x] = 0;                 // Mark x as outside.
        cnt--;                          // Decrease total count.
    };

    // Add an anonymous guest with name nm.
    auto open_anon = [&](const string& nm) {
        out.push_back("+ " + nm);        // Output enter record.
        anon.push_back(nm);              // Remember name so we can close it later.
        cnt++;                           // Increase total count.
    };

    // Remove the most recently added anonymous guest.
    auto close_anon = [&]() {
        out.push_back("- " + anon.back()); // Output leave record for top anonymous guest.
        anon.pop_back();                   // Remove it from stack.
        cnt--;                             // Decrease total count.
    };

    for(int i = 0; i < n; i++) {     // Process original records left to right.
        int t = rec_type[i], a = rec_arg[i]; // Current record type and argument.

        if(t == 0) {                 // Original record is '+ name'.
            if(present[a]) {         // If this person is already inside.
                close_named(a);      // Insert '- name' before '+ name'.
            }

            if(last_at[a] == i) {    // If this is the last original occurrence of this name.
                open_anon(names[a]); // After this, the exact identity no longer matters.
            } else {
                open_named(a);       // Otherwise keep them as a named active guest.
            }

        } else if(t == 1) {          // Original record is '- name'.
            if(present[a]) {         // If the person is inside.
                close_named(a);      // The record is valid, output it.
            } else {                 // Person is not inside.
                out.push_back("+ " + names[a]); // Insert matching enter.
                out.push_back("- " + names[a]); // Then output/perform leave.
                // Count does not change overall.
            }

        } else {                     // Original record is '= k'.
            int k = a;               // Required number of people inside.

            if(cnt < k) {            // Need to add people.
                int need = k - cnt;  // Number of enter records required.

                vector<pair<int, int>> cand; // Candidates: {next occurrence position, name id}.

                for(int x = 0; x < d; x++) { // Check all original names.
                    if(!present[x]) {        // Only absent names can be opened.
                        int nx = next_occ(x, i); // Next occurrence of x.
                        if(nx != INT_MAX && rec_type[nx] == 1) {
                            // If the next occurrence is '- x', opening x now helps later.
                            cand.push_back({nx, x});
                        }
                    }
                }

                sort(cand.begin(), cand.end()); // Prefer earliest future '- x'.

                int done = 0;        // Number of people already added.

                for(auto& pr: cand) { // Use good candidates first.
                    if(done >= need) {
                        break;       // Enough people added.
                    }
                    open_named(pr.second); // Insert '+ name'.
                    done++;
                }

                while(done < need) { // If still not enough people.
                    open_anon(next_fresh()); // Add fresh anonymous guest.
                    done++;
                }

            } else if(cnt > k) {     // Need to remove people.
                int need = cnt - k;  // Number of leave records required.

                vector<pair<int, int>> liab, useful;
                // liab: present people whose next record is '+ x';
                // useful: present people whose next record is '- x'.

                for(int x = 0; x < d; x++) { // Check all named present people.
                    if(present[x]) {
                        int nx = next_occ(x, i); // Next occurrence of this name.

                        if(nx != INT_MAX && rec_type[nx] == 0) {
                            // If next record is '+ x', x must be closed before it anyway.
                            liab.push_back({nx, x});
                        } else {
                            // Otherwise x is useful to keep, usually for a future '- x'.
                            useful.push_back({nx, x});
                        }
                    }
                }

                sort(liab.begin(), liab.end()); // Close liabilities with earliest '+ x' first.

                sort(useful.begin(), useful.end(), greater<>());
                // If forced to close useful people, close those with latest future '- x' first.

                int rem = need;      // Remaining people to remove.

                for(auto& pr: liab) { // First close liabilities.
                    if(rem == 0) {
                        break;
                    }
                    close_named(pr.second);
                    rem--;
                }

                while(rem > 0 && !anon.empty()) {
                    // Then close anonymous people, since they have no future value.
                    close_anon();
                    rem--;
                }

                for(auto& pr: useful) { // Finally close useful named people if unavoidable.
                    if(rem == 0) {
                        break;
                    }
                    close_named(pr.second);
                    rem--;
                }
            }

            out.push_back("= " + to_string(k)); // Output the original equality record.
        }
    }

    cout << out.size() << '\n';      // Output final log length.

    for(auto& line: out) {           // Output every record of the polished log.
        cout << line << '\n';
    }
}

int main() {
    ios_base::sync_with_stdio(false); // Fast input/output.
    cin.tie(nullptr);                 // Do not flush cout before cin.

    int T = 1;                        // Single test case.
    // cin >> T;                      // Multiple tests are not used in this problem.

    for(int test = 1; test <= T; test++) {
        read();                       // Read input.
        solve();                      // Solve and output answer.
    }

    return 0;                         // Successful termination.
}
```

---

## 4. Python solution with detailed comments

```python
import sys
from bisect import bisect_left


def main():
    # Read all input lines.
    data = sys.stdin.read().strip().splitlines()

    # Number of existing records.
    n = int(data[0])

    # rec_type[i]:
    #   0 means '+'
    #   1 means '-'
    #   2 means '='
    rec_type = [0] * n

    # rec_arg[i]:
    #   for '+/-': compressed name id
    #   for '=': required count
    rec_arg = [0] * n

    # Mapping from name string to integer id.
    name_id = {}

    # names[id] = original name string.
    names = []

    # Parse all records.
    for i in range(n):
        parts = data[i + 1].split()
        op = parts[0]

        if op == "=":
            # Equality record.
            rec_type[i] = 2
            rec_arg[i] = int(parts[1])
        else:
            # Enter or leave record.
            nm = parts[1]

            # Assign id to name if first time seen.
            if nm not in name_id:
                name_id[nm] = len(names)
                names.append(nm)

            rec_type[i] = 0 if op == "+" else 1
            rec_arg[i] = name_id[nm]

    # Number of distinct original names.
    d = len(names)

    # occ[x] stores all positions where name x appears in '+/-' records.
    occ = [[] for _ in range(d)]

    for i in range(n):
        if rec_type[i] != 2:
            occ[rec_arg[i]].append(i)

    # last_at[x] stores last occurrence of name x.
    last_at = [-1] * d

    for x in range(d):
        if occ[x]:
            last_at[x] = occ[x][-1]

    def next_occ(x, pos):
        """
        Return first occurrence position of name x at or after pos.
        If it does not exist, return a large value.
        """
        arr = occ[x]
        j = bisect_left(arr, pos)
        if j == len(arr):
            return 10**9
        return arr[j]

    # Original names should not be used as fresh generated names.
    used_original = set(names)

    # Base-26 counter for generating fresh names:
    # [0] -> 'a', [1] -> 'b', ..., [25] -> 'z', [0,0] -> 'aa', etc.
    digits = [0]

    def next_fresh():
        """
        Generate a lowercase name that does not occur in the original log.
        The generator itself never goes backwards, so generated names are unique.
        """
        nonlocal digits

        while True:
            # Convert digits to a string.
            s = "".join(chr(ord("a") + v) for v in digits)

            # Increment base-26 counter.
            p = len(digits) - 1

            # Carry through trailing 25s.
            while p >= 0 and digits[p] == 25:
                digits[p] = 0
                p -= 1

            if p < 0:
                # z -> aa, zz -> aaa, etc.
                digits = [0] * (len(digits) + 1)
            else:
                digits[p] += 1

            # Skip names already present in original input.
            if s not in used_original:
                return s

    # present[x] is True if original name x is currently inside as a named guest.
    present = [False] * d

    # Anonymous guests currently inside.
    # We keep their names so we can output valid leave records.
    anon = []

    # Total number of guests currently inside.
    cnt = 0

    # Output polished log.
    out = []

    def open_named(x):
        """Output '+ name' and mark original name x as inside."""
        nonlocal cnt
        out.append("+ " + names[x])
        present[x] = True
        cnt += 1

    def close_named(x):
        """Output '- name' and mark original name x as outside."""
        nonlocal cnt
        out.append("- " + names[x])
        present[x] = False
        cnt -= 1

    def open_anon(nm):
        """Output '+ nm' and remember this anonymous guest."""
        nonlocal cnt
        out.append("+ " + nm)
        anon.append(nm)
        cnt += 1

    def close_anon():
        """Close the most recently opened anonymous guest."""
        nonlocal cnt
        nm = anon.pop()
        out.append("- " + nm)
        cnt -= 1

    # Process original records in order.
    for i in range(n):
        t = rec_type[i]
        a = rec_arg[i]

        if t == 0:
            # Original record: '+ name'

            if present[a]:
                # Cannot enter twice, so insert '- name' first.
                close_named(a)

            if last_at[a] == i:
                # If this name never appears again, we do not need to track it by id.
                open_anon(names[a])
            else:
                # Otherwise keep it as a named active visitor.
                open_named(a)

        elif t == 1:
            # Original record: '- name'

            if present[a]:
                # Valid leave.
                close_named(a)
            else:
                # Person is not inside, so insert a matching enter immediately before.
                out.append("+ " + names[a])
                out.append("- " + names[a])
                # cnt does not change overall.

        else:
            # Original record: '= k'
            k = a

            if cnt < k:
                # Need to insert enter records.
                need = k - cnt

                candidates = []

                for x in range(d):
                    if not present[x]:
                        nx = next_occ(x, i)

                        # Best candidates are absent people whose next event is '- x'.
                        if nx != 10**9 and rec_type[nx] == 1:
                            candidates.append((nx, x))

                # Prefer the earliest future '- x'.
                candidates.sort()

                done = 0

                # Open useful candidates first.
                for _, x in candidates:
                    if done == need:
                        break
                    open_named(x)
                    done += 1

                # Fill remaining slots with fresh anonymous people.
                while done < need:
                    open_anon(next_fresh())
                    done += 1

            elif cnt > k:
                # Need to insert leave records.
                need = cnt - k

                liabilities = []
                useful = []

                for x in range(d):
                    if present[x]:
                        nx = next_occ(x, i)

                        if nx != 10**9 and rec_type[nx] == 0:
                            # If next event is '+ x', x must be closed before it anyway.
                            liabilities.append((nx, x))
                        else:
                            # Usually next event is '- x', so keeping x is useful.
                            useful.append((nx, x))

                # Close liabilities with earliest future '+ x' first.
                liabilities.sort()

                # If forced to close useful people, close those with latest future '- x' first.
                useful.sort(reverse=True)

                rem = need

                # First close liabilities.
                for _, x in liabilities:
                    if rem == 0:
                        break
                    close_named(x)
                    rem -= 1

                # Then close anonymous people.
                while rem > 0 and anon:
                    close_anon()
                    rem -= 1

                # Finally close useful named people if still necessary.
                for _, x in useful:
                    if rem == 0:
                        break
                    close_named(x)
                    rem -= 1

            # Now the count is exactly k, so output original equality record.
            out.append("= " + str(k))

    # Print final log length and records.
    print(len(out))
    print("\n".join(out))


if __name__ == "__main__":
    main()
```

---

## 5. Compressed editorial

Scan the log from left to right.

Maintain:

- currently present original names;
- a stack of anonymous people;
- total count.

For `+ x`:

- if `x` is already inside, insert `- x`;
- then process `+ x`;
- if this is the last occurrence of `x`, store it as anonymous.

For `- x`:

- if `x` is inside, process it;
- otherwise insert `+ x` immediately before it.

For `= k`:

- if current count is smaller, insert `k - cnt` enters.
  Prefer absent names whose next original event is `- x`, earliest such event first. They will make future leaves valid.
  Fill the rest with fresh anonymous names.

- if current count is larger, insert `cnt - k` leaves.
  First close present names whose next original event is `+ x`, earliest first, because they must be closed before that future enter anyway.
  Then close anonymous people.
  Finally, if necessary, close names whose next event is `- x`, latest first.

The immediate number of insertions around each `= k` is forced, and the greedy identity choices maximize usefulness for future records. Therefore the final log length is minimal.