A. Did We Get Everything Covered? time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output You are given two integers n and k along with a string s .

Your task is to check whether all possible strings of length n that can be formed using the first k lowercase English alphabets occur as a subsequence of s . If the answer is NO, you also need to print a string of length n that can be formed using the first k lowercase English alphabets which does not occur as a subsequence of s .

If there are multiple answers, you may print any of them.

Note: A string a is called a subsequence of another string b if a can be obtained by deleting some (possibly zero) characters from b without changing the order of the remaining characters.

Input The first line of input contains a single integer t(1≤t≤105) , the number of test cases.

The first line of each test case contains 3 integers n(1≤n≤26),k(1≤k≤26),m(1≤m≤1000) , where n and k are the same as described in the input and m is the length of the string s .

The second line of each test case contains a single string s of length m , comprising only of the first k lowercase English alphabets.

It is guaranteed that the sum of m and the sum of n over all test cases does not exceed 106 .

Output For each test case, print YES if all possible strings of length n that can be formed using the first k lowercase English alphabets occur as a subsequence of s , else print NO.

If your answer is NO, print a string of length n that can be formed using the first k lowercase English alphabets which does not occur as a subsequence of s in the next line.

You may print each letter of YES or NO in any case (for example, YES, yES, YeS will all be recognized as a positive answer).

include

include

include

using namespace std;

int main() { int t; cin >> t; while (t--) { int n, k, m; string s, t; cin >> n >> k >> m; cin >> s; int j = 0, ok = 1;

for (int i = 0; i < n; i++)
    {
        int cnt = 0;
        vector<int> check(k);

        while (cnt < k && j < m)
        {
            cnt += !check[s[j] - 'a'];
            check[s[j] - 'a'] = 1;
            j++;
        }

        if (cnt < k)
        {
            ok = 0;
            for (int i = 0; i < k; i++)
                if (!check[i])
                {
                    t += 'a' + i;
                    break;
                }
        }
        else
            t += s[j - 1];
    }

    if (ok)
        cout << "YES" << endl;
    else
    {
        cout << "NO" << endl;
        cout << t << endl;
    }
}
return 0;

}