## 1. Abridged problem statement

Jill uses one stack of dirty dishes. A lowercase letter `c` means she put a dish of color `c` onto the stack. An uppercase letter `C` means she removed a dish of color `c` from the top of the stack. Jack recorded these actions, but each `*` means he left and may have missed some actions.

The stack is empty before and after the whole process. Given Jack’s notes, determine the minimum possible number of dishes Jill cleaned in total, or print `-1` if the notes are impossible.

The string length is at most `2500`, and there are at most `5` asterisks.

---

## 2. Detailed editorial

### Stack interpretation

Lowercase letters are pushes, uppercase letters are pops. A valid complete sequence must behave like a correctly nested colored bracket sequence:

- `a` opens a dish of color `a`;
- `A` closes/removes a dish of color `a`;
- the removed dish must be on top, so colors must match in stack order.

If the completed action sequence has length `L`, then exactly `L / 2` dishes were cleaned, because every dish is pushed once and popped once.

The observed letters are fixed. Each `*` can be replaced by some unknown sequence of actions. We want to minimize the total number of actions, equivalently the number of inserted actions inside `*`.

---

### Step 1: Greedy reduction of forced observed matches

We scan the given string and build a reduced string `t`.

Whenever we see a lowercase letter or `*`, we append it.

When we see an uppercase letter:

- If the current reduced string ends with the matching lowercase letter, then this pair is forced to match, so we remove the lowercase letter.
- If the current reduced string ends with a different lowercase letter, the notes are impossible.
- Otherwise, we append the uppercase letter.

Example:

```text
afFaAA
```

Processing:

```text
a f F  -> f and F cancel
a a A  -> second a and A cancel
A      -> remains unmatched
```

The forced pairs are already valid and require no extra missing actions, but they still count as real observed actions. Therefore the solution stores the number of observed letters in `kept`.

After this reduction, we only need to determine the minimum number of missing actions needed to balance the reduced string `t`.

A useful property after reduction:

Between two asterisks, every block of ordinary letters has the form

```text
UPPERCASE...UPPERCASE lowercase...lowercase
```

because any lowercase followed later by an uppercase would either cancel or create a contradiction.

---

### Step 2: Dynamic programming on intervals

Let

```cpp
f(l, r)
```

be the minimum number of missing actions that must be generated by asterisks inside `t[l..r]` so that this substring becomes a valid balanced sequence.

If impossible, `f(l, r) = INF`.

Base case:

```cpp
f(l, r) = 0 if l > r
```

because an empty segment is already balanced.

---

### Endpoint transitions

We inspect the endpoints of the interval.

#### Both endpoints are `*`

One of them may generate nothing relevant for this interval, so:

```cpp
f(l, r) = min(f(l + 1, r), f(l, r - 1))
```

#### Left endpoint is `*`

The left `*` may be ignored:

```cpp
f(l, r) = f(l + 1, r)
```

If the right endpoint is an uppercase letter, then the left `*` can generate the matching lowercase push for it:

```cpp
f(l, r) = f(l, r - 1) + 1
```

#### Right endpoint is `*`

Symmetric case.

The right `*` may be ignored:

```cpp
f(l, r) = f(l, r - 1)
```

If the left endpoint is lowercase, then the right `*` can generate the matching uppercase pop:

```cpp
f(l, r) = f(l + 1, r) + 1
```

#### Both endpoints are observed letters

If `t[l]` is lowercase, `t[r]` is uppercase, and they have the same color, they can match:

```cpp
f(l, r) = f(l + 1, r - 1)
```

---

### Splitting the interval

A balanced sequence may be a concatenation of two balanced sequences:

```text
S = A B
```

Naively trying every split would be too slow: `O(n^3)`.

However, after greedy reduction, useful split positions are limited.

A split can only happen:

1. At an asterisk.
2. Between an uppercase run and a lowercase run, i.e. at a position `i` where:

```cpp
isupper(t[i]) && islower(t[i + 1])
```

There are at most `5` asterisks, so there are also only few such natural split positions.

For an ordinary split after position `i`:

```cpp
f(l, r) = min(f(l, r), f(l, i) + f(i + 1, r))
```

For a split at a `*`, the same asterisk can contribute actions to both sides, so it is included in both intervals:

```cpp
f(l, r) = min(f(l, r), f(l, i) + f(i, r))
```

---

### Final answer

Let:

- `kept` be the number of observed non-asterisk letters.
- `extra = f(0, n - 1)` be the minimum number of missing actions.

Then the total number of actions is:

```cpp
kept + extra
```

Each dish corresponds to one push and one pop, so the answer is:

```cpp
(kept + extra) / 2
```

If the DP says impossible, print `-1`.

Complexity:

- There are `O(n^2)` intervals.
- For each interval, only `O(number of stars)` or similar many split points are tried.
- Since there are at most `5` stars, the solution is effectively `O(n^2)`.

---

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

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

using namespace std;

// Output operator for pairs, useful for debugging.
// Not essential for the algorithm.
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

// Input operator for pairs, useful for generic reading.
// Not essential for the algorithm.
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

// Input operator for vectors.
// Reads all elements of an already-sized vector.
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

// Output operator for vectors.
// Prints elements separated by spaces.
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

// Large value representing impossible state.
const int INF = 1e9;

// Original input string.
string s;

// Reduced string after deleting forced matching observed pairs.
string t;

// Length of reduced string t.
int n;

// dp[l][r] stores f(l, r), the minimum number of inserted actions
// needed to make t[l..r] balanced.
vector<vector<int>> dp;

// Positions where it is useful to try splitting an interval.
vector<int> split_pts;

// Reads the only input string.
void read() { cin >> s; }

// Recursive memoized DP.
// Returns minimum number of missing actions needed to balance t[l..r].
int f(int l, int r) {
    // Empty substring is balanced and needs no extra actions.
    if(l > r) {
        return 0;
    }

    // Reference to memoized answer for this interval.
    int& res = dp[l][r];

    // If already computed, return it.
    if(res != -1) {
        return res;
    }

    // Initially mark as impossible.
    res = INF;

    // Case 1: both endpoints are asterisks.
    if(t[l] == '*' && t[r] == '*') {
        // Left star may contribute nothing useful to this segment.
        res = min(res, f(l + 1, r));

        // Right star may contribute nothing useful to this segment.
        res = min(res, f(l, r - 1));

    // Case 2: only the left endpoint is a star.
    } else if(t[l] == '*') {
        // If the right endpoint is an observed pop,
        // the left star can generate the corresponding push.
        if(isupper(t[r])) {
            res = min(res, f(l, r - 1) + 1);
        }

        // Or the left star may be ignored for this interval.
        res = min(res, f(l + 1, r));

    // Case 3: only the right endpoint is a star.
    } else if(t[r] == '*') {
        // If the left endpoint is an observed push,
        // the right star can generate the corresponding pop.
        if(islower(t[l])) {
            res = min(res, f(l + 1, r) + 1);
        }

        // Or the right star may be ignored for this interval.
        res = min(res, f(l, r - 1));

    // Case 4: both endpoints are observed letters.
    // They can match if the left one is lowercase,
    // the right one is uppercase, and colors are equal.
    } else if(islower(t[l]) && isupper(t[r]) && t[l] == (char)tolower(t[r])) {
        // Match the two endpoints and solve the inside.
        res = min(res, f(l + 1, r - 1));
    }

    // Try splitting this balanced sequence into two balanced parts.
    for(int i: split_pts) {
        // For a star split, the star belongs to both sides.
        // Therefore i must be strictly inside the interval.
        //
        // For a normal split after i, i can be the left boundary.
        int lo = (t[i] == '*') ? l + 1 : l;

        // Check that split point i is usable inside [l, r].
        if(i >= lo && i < r) {
            // If t[i] is a star, split as [l..i] and [i..r].
            // Otherwise split as [l..i] and [i+1..r].
            int rl = (t[i] == '*') ? i : i + 1;

            // Combine the two balanced parts.
            res = min(res, f(l, i) + f(rl, r));
        }
    }

    // Return computed result.
    return res;
}

void solve() {
    /*
        First sweep greedily cancels adjacent matched pairs.

        A lowercase 'x' followed later by uppercase 'X' with no unresolved
        obstruction between them forms a forced push-pop pair and can be
        removed from the reduced representation.

        If the current unresolved top is a lowercase letter and the next
        observed uppercase letter has a different color, then the notes are
        impossible: Jill would have tried to remove a dish of the wrong color.

        The variable kept counts all observed non-star actions. Even if a pair
        is removed from the reduced string, those actions still happened and
        still contribute to the final number of dishes.
    */

    // Reduced version of the input.
    string reduced;

    // Number of observed non-asterisk letters.
    int kept = 0;

    // Scan every character in the original notes.
    for(char c: s) {
        // Count fixed observed actions.
        if(c != '*') {
            kept++;
        }

        // A star or lowercase push cannot immediately contradict the stack,
        // so append it to the reduced string.
        if(c == '*' || islower(c)) {
            reduced.push_back(c);

        // Now c is uppercase, meaning an observed pop.
        } else if(reduced.empty() || reduced.back() == '*' || isupper(reduced.back())) {
            // If there is no unmatched lowercase immediately before it,
            // keep this uppercase in the reduced string.
            reduced.push_back(c);

        // If the previous unresolved symbol is the matching lowercase,
        // this push-pop pair is forced and can be removed.
        } else if(reduced.back() == (char)tolower(c)) {
            reduced.pop_back();

        // Otherwise the previous unresolved symbol is a lowercase letter
        // of another color, so the stack top color would be wrong.
        } else {
            cout << -1 << '\n';
            return;
        }
    }

    // Store the reduced string globally.
    t = reduced;

    // Store its length.
    n = (int)t.size();

    // Clear possible split points.
    split_pts.clear();

    // Find all useful split points.
    for(int i = 0; i < n; i++) {
        // A split may happen at a star.
        //
        // A split may also happen between an uppercase block and a lowercase
        // block, represented by position i where:
        // t[i] is uppercase and t[i + 1] is lowercase.
        if(t[i] == '*' || (i + 1 < n && isupper(t[i]) && islower(t[i + 1]))) {
            split_pts.push_back(i);
        }
    }

    // Initialize DP table with -1 meaning "not computed".
    dp.assign(n, vector<int>(n, -1));

    // Compute the minimum number of missing actions.
    // If t is empty, no extra action is needed.
    int extra = (n == 0) ? 0 : f(0, n - 1);

    // If balancing is impossible, print -1.
    if(extra >= INF) {
        cout << -1 << '\n';
    } else {
        // Total actions = observed actions + missing actions.
        // Every dish contributes exactly two actions: push and pop.
        cout << (extra + kept) / 2 << '\n';
    }
}

int main() {
    // Fast input/output.
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    // The problem has one test case.
    int T = 1;

    // Multiple tests are not used.
    // cin >> T;

    // Process the test case.
    for(int test = 1; test <= T; test++) {
        read();
        solve();
    }

    return 0;
}
```

---

## 4. Python solution with detailed comments

```python
import sys
from array import array

# Large value representing an impossible state.
INF = 10 ** 9


def solve():
    # Read Jack's notes.
    s = sys.stdin.readline().strip()

    # This will store the reduced form after deleting forced observed matches.
    reduced = []

    # Number of observed non-star letters.
    # These actions always count in the final total.
    kept = 0

    # Greedy reduction.
    for ch in s:
        # Count every observed action.
        if ch != '*':
            kept += 1

        # A star or lowercase push is appended directly.
        if ch == '*' or ('a' <= ch <= 'z'):
            reduced.append(ch)

        else:
            # ch is uppercase, meaning an observed pop.

            # If there is no lowercase push immediately available to match,
            # keep this uppercase as unresolved.
            if not reduced or reduced[-1] == '*' or ('A' <= reduced[-1] <= 'Z'):
                reduced.append(ch)

            # If the previous unresolved symbol is the matching lowercase,
            # this observed pair is forced and can be removed.
            elif reduced[-1] == ch.lower():
                reduced.pop()

            # Otherwise the top dish would have the wrong color.
            else:
                print(-1)
                return

    # Reduced string.
    t = ''.join(reduced)

    # Length of reduced string.
    n = len(t)

    # If everything was reduced away, no missing action is needed.
    if n == 0:
        print(kept // 2)
        return

    # Precompute character types for speed and clarity.
    is_low = [('a' <= c <= 'z') for c in t]
    is_up = [('A' <= c <= 'Z') for c in t]

    # lowercase version of each character, used for comparing colors.
    lower = [c.lower() for c in t]

    # Collect useful split points.
    split_pts = []

    for i in range(n):
        # Split at a star.
        if t[i] == '*':
            split_pts.append(i)

        # Split between an uppercase block and lowercase block.
        elif i + 1 < n and is_up[i] and is_low[i + 1]:
            split_pts.append(i)

    # dp[l][r] = minimum number of inserted actions needed
    # to make t[l..r] balanced.
    #
    # Using array('i') keeps memory much smaller than a list of Python ints.
    dp = [array('i', [-1]) * n for _ in range(n)]

    # Fill intervals by increasing length.
    # All transitions go to shorter intervals, so this order works.
    for length in range(1, n + 1):
        for l in range(0, n - length + 1):
            r = l + length - 1

            # Start as impossible.
            best = INF

            # Case 1: both endpoints are stars.
            if t[l] == '*' and t[r] == '*':
                # Ignore/drop left star for this segment.
                val = 0 if l + 1 > r else dp[l + 1][r]
                if val < best:
                    best = val

                # Ignore/drop right star for this segment.
                val = 0 if l > r - 1 else dp[l][r - 1]
                if val < best:
                    best = val

            # Case 2: left endpoint is a star.
            elif t[l] == '*':
                # If right endpoint is uppercase, left star can generate
                # its matching lowercase push.
                if is_up[r]:
                    val = 0 if l > r - 1 else dp[l][r - 1]
                    if val + 1 < best:
                        best = val + 1

                # Or left star contributes nothing useful here.
                val = 0 if l + 1 > r else dp[l + 1][r]
                if val < best:
                    best = val

            # Case 3: right endpoint is a star.
            elif t[r] == '*':
                # If left endpoint is lowercase, right star can generate
                # its matching uppercase pop.
                if is_low[l]:
                    val = 0 if l + 1 > r else dp[l + 1][r]
                    if val + 1 < best:
                        best = val + 1

                # Or right star contributes nothing useful here.
                val = 0 if l > r - 1 else dp[l][r - 1]
                if val < best:
                    best = val

            # Case 4: both endpoints are observed letters.
            # They may match if left is lowercase, right is uppercase,
            # and both have the same color.
            elif is_low[l] and is_up[r] and lower[l] == lower[r]:
                val = 0 if l + 1 > r - 1 else dp[l + 1][r - 1]
                if val < best:
                    best = val

            # Try all useful split points.
            for i in split_pts:
                # Star split:
                # The star can be shared by both sides,
                # so split as [l..i] and [i..r].
                if t[i] == '*':
                    if l < i < r:
                        val = dp[l][i] + dp[i][r]
                        if val < best:
                            best = val

                # Ordinary split after position i:
                # split as [l..i] and [i+1..r].
                else:
                    if l <= i < r:
                        val = dp[l][i] + dp[i + 1][r]
                        if val < best:
                            best = val

            # Store answer for interval [l, r].
            dp[l][r] = best

    # Minimum missing actions for the whole reduced string.
    extra = dp[0][n - 1]

    # If impossible, report -1.
    if extra >= INF:
        print(-1)
    else:
        # Total actions = observed actions + inserted actions.
        # Each dish has exactly one push and one pop.
        print((kept + extra) // 2)


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

---

## 5. Compressed editorial

Treat lowercase letters as colored opening brackets and uppercase letters as matching closing brackets. A valid full history is a correctly nested colored bracket sequence. The answer is half the total number of actions.

First greedily reduce the observed string. Scan left to right. Append lowercase letters and `*`. For an uppercase letter, if the current reduced string ends in the matching lowercase letter, remove that lowercase letter because this pair is forced. If it ends in a different lowercase letter, the notes are impossible. Otherwise append the uppercase letter. Count all non-star letters separately as `kept`.

Now solve on the reduced string `t`. Let `f(l, r)` be the minimum number of missing actions generated by stars inside `t[l..r]` to make it balanced.

Transitions:

- Empty interval costs `0`.
- `* ... *`: ignore either endpoint star.
- `* ... X`: the left star may create matching lowercase for uppercase `X`, cost `+1`, or be ignored.
- `x ... *`: the right star may create matching uppercase for lowercase `x`, cost `+1`, or be ignored.
- `x ... X` with same color: match endpoints and solve inside.
- Also split into two balanced parts. Useful split points are only stars and positions between an uppercase and following lowercase. At a star, split as `[l..i] + [i..r]`; otherwise split as `[l..i] + [i+1..r]`.

Because there are at most five stars, the number of useful split positions is small. The DP has `O(n^2)` states and effectively `O(n^2)` total complexity.

If `extra = f(0, n - 1)` is impossible, print `-1`; otherwise print:

```text
(kept + extra) / 2
```