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

#### Left endpoint is `*`

The left `*` may be ignored: `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:
`f(l, r) = f(l, r - 1) + 1`

#### Right endpoint is `*`

Symmetric case. The right `*` may be ignored: `f(l, r) = f(l, r - 1)`.

If the left endpoint is lowercase, then the right `*` can generate the matching uppercase pop:
`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:
`f(l, r) = f(l + 1, r - 1)`

---

### Splitting the interval

A balanced sequence may be a concatenation of two balanced sequences `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 `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`:
`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:
`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, and `extra = f(0, n - 1)` the minimum number of missing actions.

The total number of actions is `kept + extra`. Each dish corresponds to one push and one pop, so the answer is `(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. C++ Solution

```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

const int INF = 1e9;

string s;
string t;
int n;
vector<vector<int>> dp;
vector<int> split_pts;

void read() { cin >> s; }

int f(int l, int r) {
    if(l > r) {
        return 0;
    }

    int& res = dp[l][r];
    if(res != -1) {
        return res;
    }

    res = INF;

    if(t[l] == '*' && t[r] == '*') {
        res = min(res, f(l + 1, r));
        res = min(res, f(l, r - 1));
    } else if(t[l] == '*') {
        if(isupper(t[r])) {
            res = min(res, f(l, r - 1) + 1);
        }
        res = min(res, f(l + 1, r));
    } else if(t[r] == '*') {
        if(islower(t[l])) {
            res = min(res, f(l + 1, r) + 1);
        }
        res = min(res, f(l, r - 1));
    } else if(islower(t[l]) && isupper(t[r]) && t[l] == (char)tolower(t[r])) {
        res = min(res, f(l + 1, r - 1));
    }

    for(int i: split_pts) {
        int lo = (t[i] == '*') ? l + 1 : l;
        if(i >= lo && i < r) {
            int rl = (t[i] == '*') ? i : i + 1;
            res = min(res, f(l, i) + f(rl, r));
        }
    }

    return res;
}

void solve() {
    // First sweep greedily cancels adjacent matched pairs: lowercase 'x'
    // followed by uppercase 'X' is a real push-pop pair, so pop them off a
    // stack. A lowercase top against a non-matching uppercase is an outright
    // contradiction => -1. After this reduction, t consists of stars and
    // chains shaped like upper-upper-...upper-lower-lower-...lower (any pops
    // that survive must precede any pushes that survive, otherwise they'd
    // have cancelled). Only stars can balance what remains.
    //
    // f(l, r) = min number of extra letters the stars in t[l..r] must produce
    // so that t[l..r] becomes a fully balanced bracket sequence (lowercase =
    // open, uppercase = close). Endpoint cases handle stars at the borders
    // absorbing one neighbor as their matching bracket, and a matched pair at
    // the borders. We also try every "natural split" point: a star (which can
    // split into two stars covering both halves) or a position between an
    // uppercase and a lowercase (the only place an outer balanced sequence
    // can be cut in two without crossing an active matching pair).

    string reduced;
    int kept = 0;
    for(char c: s) {
        if(c != '*') {
            kept++;
        }

        if(c == '*' || islower(c)) {
            reduced.push_back(c);
        } else if(reduced.empty() || reduced.back() == '*' || isupper(reduced.back())) {
            reduced.push_back(c);
        } else if(reduced.back() == (char)tolower(c)) {
            reduced.pop_back();
        } else {
            cout << -1 << '\n';
            return;
        }
    }

    t = reduced;
    n = (int)t.size();

    split_pts.clear();
    for(int i = 0; i < n; i++) {
        if(t[i] == '*' || (i + 1 < n && isupper(t[i]) && islower(t[i + 1]))) {
            split_pts.push_back(i);
        }
    }

    dp.assign(n, vector<int>(n, -1));

    int extra = (n == 0) ? 0 : f(0, n - 1);
    if(extra >= INF) {
        cout << -1 << '\n';
    } else {
        cout << (extra + kept) / 2 << '\n';
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    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
```
