description: You are given an integer k and a string s that consists only of characters 'a' (a lowercase Latin letter) and '*' (an asterisk).
  
  Each asterisk should be replaced with several (from 0 to k inclusive) lowercase Latin letters 'b'. Different asterisk can be replaced with different counts of letter 'b'.
  
  The result of the replacement is called a BA-string.
  
  Two strings a and b are different if they either have different lengths or there exists such a position i that a_i ≠ b_i.
  
  A string a is lexicographically smaller than a string b if and only if one of the following holds: 
  
    * a is a prefix of b, but a ≠ b; 
    * in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b. 
  
  
  
  Now consider all different BA-strings and find the x-th lexicographically smallest of them.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 2000) — the number of testcases.
  
  The first line of each testcase contains three integers n, k and x (1 ≤ n ≤ 2000; 0 ≤ k ≤ 2000; 1 ≤ x ≤ 10^{18}). n is the length of string s.
  
  The second line of each testcase is a string s. It consists of n characters, each of them is either 'a' (a lowercase Latin letter) or '*' (an asterisk).
  
  The sum of n over all testcases doesn't exceed 2000. For each testcase x doesn't exceed the total number of different BA-strings. String s contains at least one character 'a'.
  
  Output
  
  For each testcase, print a single string, consisting only of characters 'b' and 'a' (lowercase Latin letters) — the x-th lexicographically smallest BA-string.
  
  Example
  
  Input
  
  
  3
  2 4 3
  a*
  4 1 3
  a**a
  6 3 20
  **a***
  
  
  Output
  
  
  abb
  abba
  babbbbbbbbb
  
  Note
  
  In the first testcase of the example, BA-strings ordered lexicographically are: 
  
    1. a
    2. ab
    3. abb
    4. abbb
    5. abbbb
  
  
  
  In the second testcase of the example, BA-strings ordered lexicographically are: 
  
    1. aa
    2. aba
    3. abba
  
  
  
  Note that string "aba" is only counted once, even though there are two ways to replace asterisks with characters 'b' to get it.
solution: #include <bits/stdc++.h>
  using namespace std;
  const long long N = 2e3 + 7;
  long long n;
  char a[N];
  long long k;
  long long x;
  long long mim[N];
  long long ban[N];
  long long s[N];
  long long xx = 0;
  long long gon;
  void solve() {
    long long cnt = 0;
    string res = "";
    cin >> n >> k >> x;
    long long dak = 1;
    xx++;
    for (long long i = 1; i <= n; i++) {
      cin >> a[i];
      ban[i] = 0;
      mim[i] = 0;
    }
    for (long long i = 1; i <= n; i++) {
      if (a[i] == '*') {
        long long dem = 0;
        long long j = i;
        while (j <= n && a[j] == '*') {
          dem++;
          j++;
        }
        mim[++cnt] = (j - i) * k + 1;
        i = j - 1;
      }
    }
    long long pos = cnt + 1;
    for (long long i = cnt; i >= 1; i--) {
      if (dak >= x / mim[i] + bool(x % mim[i])) {
        pos = i;
        break;
      }
      dak = dak * mim[i];
      s[i] = dak;
    }
    s[cnt + 1] = 1;
    for (long long i = pos; i <= cnt; i++) {
      ban[i] = x / s[i + 1] + bool(x % s[i + 1]) - 1;
      x = x % s[i + 1];
      if (x == 0) {
        x = s[i + 1];
      }
    }
    long long id = 0;
    for (long long i = 1; i <= n; i++) {
      if (a[i] == 'a') {
        cout << 'a';
      } else {
        long long j = i;
        while (j <= n && a[j] == '*') {
          j++;
        }
        id++;
        for (long long t = 1; t <= ban[id]; t++) {
          cout << 'b';
        }
        i = j - 1;
      }
    }
    cout << '\n';
  }
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    long long test;
    cin >> test;
    gon = test;
    while (test--) {
      solve();
    }
  }
  