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 mod = 1e9 + 7;
  const long long inf = 1e17;
  long long mod_add(long long a, long long b) {
    a = a % mod;
    b = b % mod;
    return (((a + b) % mod) + mod) % mod;
  }
  long long mod_mul(long long a, long long b) {
    a = a % mod;
    b = b % mod;
    return (((a * b) % mod) + mod) % mod;
  }
  long long inv(long long i) {
    if (i == 1) return 1;
    return (mod - ((mod / i) * inv(mod % i)) % mod) % mod;
  }
  long long pwr(long long a, long long b) {
    a %= mod;
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = res * a % mod;
      a = a * a % mod;
      b >>= 1;
    }
    return res;
  }
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  string yans = "YES", nans = "NO";
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long t = 1;
    cin >> t;
    while (t--) {
      long long n, k, x;
      cin >> n >> k >> x;
      string s;
      cin >> s;
      vector<long long> v;
      string ans = "";
      long long cnt = 0;
      for (long long i = 0; i < n; i++) {
        if (s[i] == '*') {
          if (cnt == 0) {
            ans += '?';
          }
          cnt++;
        } else {
          ans += 'a';
          if (cnt) {
            v.push_back(cnt * k + 1);
            cnt = 0;
          }
        }
      }
      if (cnt) {
        v.push_back(cnt * k + 1);
      }
      long long m = v.size(), b = ans.size();
      long long j = m - 1;
      for (long long i = b - 1; i >= 0; i--) {
        if (ans[i] == '?' and x) {
          long long p = x % v[j], q = v[j];
          if (p) {
            v[j] = p;
          }
          x = (x + q - 1) / q;
          j--;
        }
      }
      j = 0;
      string res = "";
      for (long long i = 0; i < b; i++) {
        if (ans[i] == '?') {
          for (long long l = 0; l < v[j] - 1; l++) {
            res += 'b';
          }
          j++;
        } else {
          res += 'a';
        }
      }
      cout << res << "\n";
    }
    return 0;
  }
  