## 1. Abridged problem statement

Given a regular expression-like string of length at most `255` containing lowercase letters and special symbols:

- `?` must be replaced by exactly one lowercase letter.
- `!` must be replaced by exactly three lowercase letters.
- `*` may be replaced by any string of lowercase letters, including the empty string.

Determine whether the expression can be transformed into a palindrome. If yes, output `YES` and the shortest possible palindrome. If several shortest palindromes exist, output the lexicographically smallest one. Otherwise output `NO`.

---

## 2. Detailed editorial

### Step 1: Simplify `!`

The symbol `!` means “exactly three arbitrary letters”, which is equivalent to three `?` symbols.

So we first expand:

```text
! -> ???
```

After this, the string contains only:

- lowercase letters,
- `?`,
- `*`.

The original length is at most `255`, so after expansion the length is at most `765`.

Let the expanded string be `s`, with length `n`.

---

### Step 2: Dynamic programming for minimum palindrome length

Define:

```text
dp[l][r] = minimum length of a palindrome that can be produced
           from the substring s[l...r]
```

If it is impossible, `dp[l][r] = INF`.

For an empty substring:

```text
dp[l][r] = 0 if l > r
```

For one character:

- if it is `*`, it can become empty, so length `0`;
- otherwise, it must become one character, so length `1`.

---

### Transitions

We want the generated string to be a palindrome. Therefore, the first and last generated characters must be equal.

Consider substring `s[l...r]`.

#### Case 1: left side is `*`

The `*` can become empty:

```text
dp[l][r] = min(dp[l][r], dp[l + 1][r])
```

#### Case 2: right side is `*`

Similarly:

```text
dp[l][r] = min(dp[l][r], dp[l][r - 1])
```

#### Case 3: both ends are not `*`

Then both ends must generate one character each.

They are compatible if:

- both are equal letters;
- or at least one of them is `?`.

Then:

```text
dp[l][r] = min(dp[l][r], dp[l + 1][r - 1] + 2)
```

#### Case 4: left is not `*`, right is `*`

The right `*` may generate a character matching `s[l]`.

The remaining part still includes the same right `*`, because it may generate more characters.

```text
dp[l][r] = min(dp[l][r], dp[l + 1][r] + 2)
```

#### Case 5: left is `*`, right is not `*`

Symmetric:

```text
dp[l][r] = min(dp[l][r], dp[l][r - 1] + 2)
```

---

### Step 3: Reconstruct lexicographically smallest answer

The DP only tells us the shortest length. We also need the lexicographically smallest palindrome of that length.

For every interval `s[l...r]`, among all transitions that achieve `dp[l][r]`, we build the candidate string and take the lexicographically smallest one.

For example, if both ends are non-`*`, we create:

```text
c + best[l + 1][r - 1] + c
```

where `c` is the smallest possible matching character.

Examples:

- `a` and `a` force `c = 'a'`;
- `?` and `b` force `c = 'b'`;
- `?` and `?` allow any letter, so choose `c = 'a'`.

For `*`, we also consider transitions where the star becomes empty.

Because interval length increases one by one, each state depends only on shorter intervals. The C++ solution stores only three rolling layers of actual strings:

- length `len - 2`,
- length `len - 1`,
- current length `len`.

The integer DP table is kept fully.

---

### Correctness argument

The DP considers every possible way the outermost characters of a palindrome can be generated.

- If a boundary is `*`, it can either generate nothing or generate an outer character.
- If a boundary is not `*`, it must generate exactly one character.
- For a palindrome, the generated outer characters must match.

Therefore, every valid palindrome generation corresponds to one of the transitions.

The DP minimizes length, so `dp[0][n - 1]` gives the shortest possible palindrome length.

During reconstruction, we examine exactly those transitions that achieve the optimal length and choose the smallest resulting string lexicographically. Since smaller subproblems are already reconstructed optimally, each built candidate is the best possible for that transition. Taking the minimum among them gives the globally lexicographically smallest shortest palindrome.

---

### Complexity

Let `n` be the expanded length, `n <= 765`.

The DP has `O(n^2)` states and constant transitions per state.

String reconstruction may compare and concatenate strings of length up to `O(n)`, so the worst-case time is approximately `O(n^3)`.

Memory usage:

- `O(n^2)` for `dp`;
- rolling layers of strings use roughly `O(n^2)` total characters.

---

## 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;
};

string regex_str;

void read() { cin >> regex_str; }

void solve() {
    // First, '!' is just shorthand for "???", so we expand it and only have to
    // reason about letters, '?' and '*'. The core DP is dp[l][r], the minimum
    // length of a palindrome that the regex segment s[l..r] can produce. Since
    // the produced string is a palindrome, its first and last characters must
    // agree, and they come from s[l] and s[r] respectively (or, if either is a
    // '*', from that '*' absorbing one extra character). That gives a small
    // case split on (s[l], s[r]): when both are non-'*' we peel both ends with
    // a compatible outer char (letters force it, '?' is free); when one is
    // '*' and the other isn't, the '*' may match empty (drop and recurse) or
    // absorb the matching char on its side while staying in place; when both
    // are '*' either may match empty.
    //
    // Length-optimal moves can produce different palindromes, so for the
    // lex-smallest answer we have to actually compare the candidates rather
    // than rely on a heuristic ('prefer drop' / 'prefer peel' both have
    // counterexamples, e.g. "*bab*ba" needs drop but "*!ba*!" needs peel).
    // We therefore build the actual pal[l][r] string for every cell, in
    // increasing-length order. Each cell of length L depends only on cells
    // of length L-1 and L-2, so we keep just three rolling layers of strings
    // alive at any time, which caps memory at O(n^2) chars and stays well
    // under the limit even at n=765 after '!' expansion.

    string s;
    for(char c: regex_str) {
        if(c == '!') {
            s += "???";
        } else {
            s += c;
        }
    }

    int n = s.size();
    if(n == 0) {
        cout << "YES\n\n";
        return;
    }

    const int INF = INT_MAX / 2;
    vector<vector<int>> dp(n, vector<int>(n, INF));

    auto is_star = [&](int p) { return s[p] == '*'; };
    auto is_q = [&](int p) { return s[p] == '?'; };

    auto get_dp = [&](int a, int b) -> int {
        if(a > b) {
            return 0;
        }
        return dp[a][b];
    };

    for(int i = 0; i < n; i++) {
        dp[i][i] = is_star(i) ? 0 : 1;
    }

    for(int len = 2; len <= n; len++) {
        for(int l = 0; l + len - 1 < n; l++) {
            int r = l + len - 1;
            int best = INF;

            if(is_star(l) && get_dp(l + 1, r) < INF) {
                best = min(best, get_dp(l + 1, r));
            }

            if(is_star(r) && get_dp(l, r - 1) < INF) {
                best = min(best, get_dp(l, r - 1));
            }

            if(!is_star(l) && !is_star(r)) {
                int inner = get_dp(l + 1, r - 1);
                bool ok = true;
                if(!is_q(l) && !is_q(r) && s[l] != s[r]) {
                    ok = false;
                }
                if(ok && inner < INF) {
                    best = min(best, inner + 2);
                }
            } else if(!is_star(l) && is_star(r)) {
                int inner = get_dp(l + 1, r);
                if(inner < INF) {
                    best = min(best, inner + 2);
                }
            } else if(is_star(l) && !is_star(r)) {
                int inner = get_dp(l, r - 1);
                if(inner < INF) {
                    best = min(best, inner + 2);
                }
            }

            dp[l][r] = best;
        }
    }

    if(dp[0][n - 1] >= INF) {
        cout << "NO\n";
        return;
    }

    vector<string> layer_a(n), layer_b(n), layer_c(n);
    vector<string>* lm2 = &layer_a;
    vector<string>* lm1 = &layer_b;
    vector<string>* cur = &layer_c;

    string answer;
    for(int len = 1; len <= n; len++) {
        for(int l = 0; l + len - 1 < n; l++) {
            int r = l + len - 1;
            (*cur)[l].clear();
            if(dp[l][r] >= INF) {
                continue;
            }

            if(len == 1) {
                if(is_star(l)) {
                    (*cur)[l] = "";
                } else if(is_q(l)) {
                    (*cur)[l] = "a";
                } else {
                    (*cur)[l] = string(1, s[l]);
                }
                continue;
            }

            int target = dp[l][r];
            bool have = false;
            string best;

            auto try_cand = [&](string cand) {
                if(!have || cand < best) {
                    best = std::move(cand);
                    have = true;
                }
            };

            if(!is_star(l) && !is_star(r)) {
                int inner = (len == 2) ? 0 : dp[l + 1][r - 1];
                if(inner != INF && inner + 2 == target) {
                    bool ok = true;
                    char c = 0;
                    if(!is_q(l) && !is_q(r)) {
                        if(s[l] == s[r]) {
                            c = s[l];
                        } else {
                            ok = false;
                        }
                    } else if(!is_q(l)) {
                        c = s[l];
                    } else if(!is_q(r)) {
                        c = s[r];
                    } else {
                        c = 'a';
                    }
                    if(ok) {
                        const string& ip =
                            (len == 2) ? string() : (*lm2)[l + 1];
                        string cand;
                        cand.reserve(ip.size() + 2);
                        cand += c;
                        cand += ip;
                        cand += c;
                        try_cand(std::move(cand));
                    }
                }
            }

            if(!is_star(l) && is_star(r)) {
                int inner = dp[l + 1][r];
                if(inner != INF && inner + 2 == target) {
                    char c = is_q(l) ? 'a' : s[l];
                    const string& ip = (*lm1)[l + 1];
                    string cand;
                    cand.reserve(ip.size() + 2);
                    cand += c;
                    cand += ip;
                    cand += c;
                    try_cand(std::move(cand));
                }
            }

            if(is_star(l) && !is_star(r)) {
                int inner = dp[l][r - 1];
                if(inner != INF && inner + 2 == target) {
                    char c = is_q(r) ? 'a' : s[r];
                    const string& ip = (*lm1)[l];
                    string cand;
                    cand.reserve(ip.size() + 2);
                    cand += c;
                    cand += ip;
                    cand += c;
                    try_cand(std::move(cand));
                }
            }

            if(is_star(l) && dp[l + 1][r] == target) {
                try_cand((*lm1)[l + 1]);
            }

            if(is_star(r) && dp[l][r - 1] == target) {
                try_cand((*lm1)[l]);
            }

            (*cur)[l] = std::move(best);
        }

        if(len == n) {
            answer = std::move((*cur)[0]);
            break;
        }

        vector<string>* tmp = lm2;
        lm2 = lm1;
        lm1 = cur;
        cur = tmp;
        for(auto& x: *cur) {
            x.clear();
            x.shrink_to_fit();
        }
    }

    cout << "YES\n" << answer << "\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


def solve():
    # Read the regular expression.
    regex = sys.stdin.readline().strip()

    # Expand '!' into '???', because '!' means exactly three arbitrary letters.
    s = []
    for ch in regex:
        if ch == "!":
            s.append("???")
        else:
            s.append(ch)

    # Join pieces into the expanded expression.
    s = "".join(s)

    # Length after expansion.
    n = len(s)

    # Empty expression can produce the empty palindrome.
    if n == 0:
        print("YES")
        print()
        return

    # Large value representing impossible states.
    INF = 10**9

    # dp[l][r] = minimum length of palindrome obtainable from s[l:r+1].
    dp = [[INF] * n for _ in range(n)]

    # Helper function: is s[i] a star?
    def is_star(i):
        return s[i] == "*"

    # Helper function: is s[i] a question mark?
    def is_q(i):
        return s[i] == "?"

    # Helper function for DP intervals.
    # Empty interval has answer 0.
    def get_dp(l, r):
        if l > r:
            return 0
        return dp[l][r]

    # Initialize intervals of length 1.
    for i in range(n):
        if is_star(i):
            # '*' may become empty.
            dp[i][i] = 0
        else:
            # A letter or '?' must become exactly one character.
            dp[i][i] = 1

    # Fill DP by increasing interval length.
    for length in range(2, n + 1):
        for l in range(0, n - length + 1):
            r = l + length - 1
            best = INF

            # Left '*' becomes empty.
            if is_star(l) and get_dp(l + 1, r) < INF:
                best = min(best, get_dp(l + 1, r))

            # Right '*' becomes empty.
            if is_star(r) and get_dp(l, r - 1) < INF:
                best = min(best, get_dp(l, r - 1))

            # Both ends are not '*'.
            if not is_star(l) and not is_star(r):
                inner = get_dp(l + 1, r - 1)

                # Boundary characters are compatible unless both are fixed and different.
                ok = True
                if not is_q(l) and not is_q(r) and s[l] != s[r]:
                    ok = False

                if ok and inner < INF:
                    best = min(best, inner + 2)

            # Left is not '*', right is '*'.
            elif not is_star(l) and is_star(r):
                inner = get_dp(l + 1, r)

                # Right '*' generates a matching outer character.
                if inner < INF:
                    best = min(best, inner + 2)

            # Left is '*', right is not '*'.
            elif is_star(l) and not is_star(r):
                inner = get_dp(l, r - 1)

                # Left '*' generates a matching outer character.
                if inner < INF:
                    best = min(best, inner + 2)

            # Store the best result for this interval.
            dp[l][r] = best

    # If full expression cannot produce a palindrome.
    if dp[0][n - 1] >= INF:
        print("NO")
        return

    # Rolling layers of reconstructed strings.
    # lm2 stores interval length length - 2.
    # lm1 stores interval length length - 1.
    lm2 = [""] * n
    lm1 = [""] * n

    answer = ""

    # Reconstruct lexicographically smallest optimal palindrome.
    for length in range(1, n + 1):
        # Current layer for this interval length.
        cur = [""] * n

        for l in range(0, n - length + 1):
            r = l + length - 1

            # Impossible interval.
            if dp[l][r] >= INF:
                continue

            # Base case: one symbol.
            if length == 1:
                if is_star(l):
                    # '*' can produce empty.
                    cur[l] = ""
                elif is_q(l):
                    # '?' should become smallest possible letter.
                    cur[l] = "a"
                else:
                    # Fixed letter.
                    cur[l] = s[l]
                continue

            # Required optimal length.
            target = dp[l][r]

            # Best candidate string found so far.
            best = None

            # Helper to update best candidate.
            def try_candidate(cand):
                nonlocal best
                if best is None or cand < best:
                    best = cand

            # Case 1: both ends are not '*'.
            if not is_star(l) and not is_star(r):
                inner_len = 0 if length == 2 else dp[l + 1][r - 1]

                # This transition must achieve the optimal length.
                if inner_len != INF and inner_len + 2 == target:
                    ok = True
                    c = ""

                    if not is_q(l) and not is_q(r):
                        # Both are fixed letters.
                        if s[l] == s[r]:
                            c = s[l]
                        else:
                            ok = False
                    elif not is_q(l):
                        # Left fixed, right '?'.
                        c = s[l]
                    elif not is_q(r):
                        # Right fixed, left '?'.
                        c = s[r]
                    else:
                        # Both '?'.
                        c = "a"

                    if ok:
                        # Inner palindrome.
                        inner_pal = "" if length == 2 else lm2[l + 1]

                        # Build candidate.
                        cand = c + inner_pal + c

                        # Consider candidate.
                        try_candidate(cand)

            # Case 2: left is not '*', right is '*'.
            if not is_star(l) and is_star(r):
                inner_len = dp[l + 1][r]

                # Right '*' generates the matching outer character.
                if inner_len != INF and inner_len + 2 == target:
                    c = "a" if is_q(l) else s[l]
                    inner_pal = lm1[l + 1]
                    cand = c + inner_pal + c
                    try_candidate(cand)

            # Case 3: left is '*', right is not '*'.
            if is_star(l) and not is_star(r):
                inner_len = dp[l][r - 1]

                # Left '*' generates the matching outer character.
                if inner_len != INF and inner_len + 2 == target:
                    c = "a" if is_q(r) else s[r]
                    inner_pal = lm1[l]
                    cand = c + inner_pal + c
                    try_candidate(cand)

            # Case 4: left '*' becomes empty.
            if is_star(l) and dp[l + 1][r] == target:
                try_candidate(lm1[l + 1])

            # Case 5: right '*' becomes empty.
            if is_star(r) and dp[l][r - 1] == target:
                try_candidate(lm1[l])

            # Store best candidate for current interval.
            cur[l] = best if best is not None else ""

        # Full interval reconstructed.
        if length == n:
            answer = cur[0]
            break

        # Rotate layers.
        lm2, lm1 = lm1, cur

    # Output result.
    print("YES")
    print(answer)


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

---

## 5. Compressed editorial

Expand every `!` into `???`. Let the resulting string be `s`.

Use interval DP:

```text
dp[l][r] = shortest palindrome length obtainable from s[l...r]
```

Base:

```text
dp[i][i] = 0 if s[i] == '*', else 1
dp[empty] = 0
```

Transitions:

- left `*` empty:

```text
dp[l][r] = min(dp[l][r], dp[l + 1][r])
```

- right `*` empty:

```text
dp[l][r] = min(dp[l][r], dp[l][r - 1])
```

- both ends non-`*`, compatible:

```text
dp[l][r] = min(dp[l][r], dp[l + 1][r - 1] + 2)
```

- left non-`*`, right `*`:

```text
dp[l][r] = min(dp[l][r], dp[l + 1][r] + 2)
```

- left `*`, right non-`*`:

```text
dp[l][r] = min(dp[l][r], dp[l][r - 1] + 2)
```

If `dp[0][n - 1]` is impossible, print `NO`.

Otherwise reconstruct the answer by increasing interval length. For each interval, try every transition that gives the optimal length and choose the lexicographically smallest resulting palindrome. For `?`, use the smallest possible matching character, usually `'a'` unless forced by the opposite side.

Complexity is `O(n^2)` DP states, with string reconstruction up to `O(n^3)` time in the worst case.