## 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. Provided C++ solution with detailed comments

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

using namespace std; // Allows using standard library names without std::.

// Output operator for pairs, useful for debugging or generic printing.
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second; // Print first and second separated by space.
}

// Input operator for pairs.
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second; // Read first and second.
}

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

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

string regex_str; // Stores the original input expression.

void read() { cin >> regex_str; } // Reads the input expression.

// Main solving function.
void solve() {
    // Expand '!' into "???", because '!' means exactly three arbitrary letters.
    string s; // Expanded expression.
    for(char c: regex_str) { // Process every character of the original expression.
        if(c == '!') { // If the character is '!'.
            s += "???"; // Replace it by three '?' characters.
        } else {
            s += c; // Otherwise keep the character unchanged.
        }
    }

    int n = s.size(); // Length of the expanded expression.

    if(n == 0) { // Empty expression can produce the empty palindrome.
        cout << "YES\n\n"; // Output YES and an empty second line.
        return; // Finish.
    }

    const int INF = INT_MAX / 2; // Large value representing impossible state.

    // dp[l][r] = minimum palindrome length obtainable from s[l...r].
    vector<vector<int>> dp(n, vector<int>(n, INF));

    // Helper lambda: checks whether position p is '*'.
    auto is_star = [&](int p) { return s[p] == '*'; };

    // Helper lambda: checks whether position p is '?'.
    auto is_q = [&](int p) { return s[p] == '?'; };

    // Helper lambda to safely get dp value.
    // Empty interval has minimum palindrome length 0.
    auto get_dp = [&](int a, int b) -> int {
        if(a > b) { // Empty substring.
            return 0;
        }
        return dp[a][b]; // Normal substring.
    };

    // Initialize intervals of length 1.
    for(int i = 0; i < n; i++) {
        // '*' may become empty, so length 0.
        // Any other symbol must generate exactly one character.
        dp[i][i] = is_star(i) ? 0 : 1;
    }

    // Compute DP for all interval lengths from 2 to n.
    for(int len = 2; len <= n; len++) {
        // Iterate over all left endpoints.
        for(int l = 0; l + len - 1 < n; l++) {
            int r = l + len - 1; // Right endpoint of current interval.
            int best = INF; // Best answer for dp[l][r].

            // If left character is '*', it may generate the empty string.
            if(is_star(l) && get_dp(l + 1, r) < INF) {
                best = min(best, get_dp(l + 1, r));
            }

            // If right character is '*', it may generate the empty string.
            if(is_star(r) && get_dp(l, r - 1) < INF) {
                best = min(best, get_dp(l, r - 1));
            }

            // If both ends are not '*', both generate exactly one character.
            if(!is_star(l) && !is_star(r)) {
                int inner = get_dp(l + 1, r - 1); // Minimum length of inner palindrome.
                bool ok = true; // Whether the two boundary characters can match.

                // If both are fixed letters and they differ, they cannot match.
                if(!is_q(l) && !is_q(r) && s[l] != s[r]) {
                    ok = false;
                }

                // If compatible and the inner interval is possible.
                if(ok && inner < INF) {
                    best = min(best, inner + 2); // Add two matching outer characters.
                }
            }
            // Left is not '*', right is '*'.
            // The right '*' may generate a character equal to the left boundary.
            else if(!is_star(l) && is_star(r)) {
                int inner = get_dp(l + 1, r); // Right star remains for the inner part.
                if(inner < INF) {
                    best = min(best, inner + 2); // Add matching outer pair.
                }
            }
            // Left is '*', right is not '*'.
            // The left '*' may generate a character equal to the right boundary.
            else if(is_star(l) && !is_star(r)) {
                int inner = get_dp(l, r - 1); // Left star remains for the inner part.
                if(inner < INF) {
                    best = min(best, inner + 2); // Add matching outer pair.
                }
            }

            dp[l][r] = best; // Store the best value for this interval.
        }
    }

    // If the whole expression cannot generate a palindrome.
    if(dp[0][n - 1] >= INF) {
        cout << "NO\n"; // Output NO.
        return; // Finish.
    }

    // These three arrays store reconstructed strings for rolling interval lengths.
    vector<string> layer_a(n), layer_b(n), layer_c(n);

    // lm2 points to strings for intervals of length len - 2.
    vector<string>* lm2 = &layer_a;

    // lm1 points to strings for intervals of length len - 1.
    vector<string>* lm1 = &layer_b;

    // cur points to strings for intervals of current length len.
    vector<string>* cur = &layer_c;

    string answer; // Final reconstructed answer.

    // Reconstruct strings by increasing interval length.
    for(int len = 1; len <= n; len++) {
        // Iterate over all intervals of this length.
        for(int l = 0; l + len - 1 < n; l++) {
            int r = l + len - 1; // Right endpoint.

            (*cur)[l].clear(); // Clear current stored string.

            // If this interval cannot produce a palindrome, leave it empty.
            if(dp[l][r] >= INF) {
                continue;
            }

            // Base case: interval of length 1.
            if(len == 1) {
                if(is_star(l)) {
                    (*cur)[l] = ""; // '*' can produce empty string.
                } else if(is_q(l)) {
                    (*cur)[l] = "a"; // '?' should be smallest possible letter.
                } else {
                    (*cur)[l] = string(1, s[l]); // Fixed letter.
                }
                continue; // Done with this interval.
            }

            int target = dp[l][r]; // Required optimal length for this interval.
            bool have = false; // Whether we already have a candidate.
            string best; // Best lexicographic candidate.

            // Tries to improve the current best string.
            auto try_cand = [&](string cand) {
                if(!have || cand < best) { // First candidate or lexicographically smaller.
                    best = std::move(cand); // Store candidate efficiently.
                    have = true; // Mark that we have a candidate.
                }
            };

            // Case: both ends are not '*'.
            if(!is_star(l) && !is_star(r)) {
                // Inner optimal length.
                int inner = (len == 2) ? 0 : dp[l + 1][r - 1];

                // This transition is valid only if it gives the optimal target.
                if(inner != INF && inner + 2 == target) {
                    bool ok = true; // Whether boundary characters can match.
                    char c = 0; // The chosen outer character.

                    if(!is_q(l) && !is_q(r)) {
                        // Both are fixed letters.
                        if(s[l] == s[r]) {
                            c = s[l]; // They must be equal.
                        } else {
                            ok = false; // Different fixed letters cannot match.
                        }
                    } else if(!is_q(l)) {
                        c = s[l]; // Left is fixed, right '?' becomes same.
                    } else if(!is_q(r)) {
                        c = s[r]; // Right is fixed, left '?' becomes same.
                    } else {
                        c = 'a'; // Both are '?', choose lexicographically smallest.
                    }

                    if(ok) {
                        // Inner palindrome string.
                        const string& ip =
                            (len == 2) ? string() : (*lm2)[l + 1];

                        string cand; // Candidate palindrome.
                        cand.reserve(ip.size() + 2); // Reserve memory.
                        cand += c; // Add left outer character.
                        cand += ip; // Add inner palindrome.
                        cand += c; // Add right outer character.
                        try_cand(std::move(cand)); // Consider this candidate.
                    }
                }
            }

            // Case: left is not '*', right is '*'.
            if(!is_star(l) && is_star(r)) {
                int inner = dp[l + 1][r]; // Right '*' remains in inner interval.

                // Valid only if this transition achieves optimal length.
                if(inner != INF && inner + 2 == target) {
                    char c = is_q(l) ? 'a' : s[l]; // Choose smallest possible left char.
                    const string& ip = (*lm1)[l + 1]; // Inner palindrome.

                    string cand; // Candidate string.
                    cand.reserve(ip.size() + 2); // Reserve needed size.
                    cand += c; // Outer left character.
                    cand += ip; // Inner palindrome.
                    cand += c; // Matching character generated by right '*'.
                    try_cand(std::move(cand)); // Consider candidate.
                }
            }

            // Case: left is '*', right is not '*'.
            if(is_star(l) && !is_star(r)) {
                int inner = dp[l][r - 1]; // Left '*' remains in inner interval.

                // Valid only if this transition achieves optimal length.
                if(inner != INF && inner + 2 == target) {
                    char c = is_q(r) ? 'a' : s[r]; // Choose smallest possible right char.
                    const string& ip = (*lm1)[l]; // Inner palindrome.

                    string cand; // Candidate string.
                    cand.reserve(ip.size() + 2); // Reserve needed memory.
                    cand += c; // Character generated by left '*'.
                    cand += ip; // Inner palindrome.
                    cand += c; // Right outer character.
                    try_cand(std::move(cand)); // Consider candidate.
                }
            }

            // Case: left '*' becomes empty.
            if(is_star(l) && dp[l + 1][r] == target) {
                try_cand((*lm1)[l + 1]); // Candidate is the interval without left '*'.
            }

            // Case: right '*' becomes empty.
            if(is_star(r) && dp[l][r - 1] == target) {
                try_cand((*lm1)[l]); // Candidate is the interval without right '*'.
            }

            (*cur)[l] = std::move(best); // Store best string for interval [l, r].
        }

        // If we just computed the whole interval, save final answer.
        if(len == n) {
            answer = std::move((*cur)[0]); // Whole answer is at interval [0, n-1].
            break; // Reconstruction complete.
        }

        // Rotate layers:
        // old len - 1 becomes len - 2,
        // current becomes len - 1,
        // old len - 2 buffer becomes reusable current buffer.
        vector<string>* tmp = lm2;
        lm2 = lm1;
        lm1 = cur;
        cur = tmp;

        // Clear the reusable buffer to free memory held by old strings.
        for(auto& x: *cur) {
            x.clear(); // Make string empty.
            x.shrink_to_fit(); // Release unused capacity if possible.
        }
    }

    // Output successful answer.
    cout << "YES\n" << answer << "\n";
}

int main() {
    ios_base::sync_with_stdio(false); // Speed up C++ IO.
    cin.tie(nullptr); // Untie cin from cout for faster input.

    int T = 1; // Number of tests. Problem has one test.

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

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

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

---

## 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.