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;
  long long mod = 1e9 + 7;
  long long INF = 1e18;
  bool isPrime(long long n) {
    if (n <= 1) return false;
    if (n <= 3) return true;
    if (n % 2 == 0 || n % 3 == 0) return false;
    for (long long i = 5; i * i <= n; i = i + 6)
      if (n % i == 0 || n % (i + 2) == 0) return false;
    return true;
  }
  long long nCr(long long n, long long r) {
    long long res = 1;
    if (n < r) return -1;
    r = min(r, n - r);
    for (long long i = 0; i < r; i++) {
      res *= (n - i);
      res /= (i + 1);
    }
    return res;
  }
  long long nPr(long long n, long long r) {
    long long res = 1;
    if (n < r) return -1;
    for (long long i = 0; i < r; i++) res *= (n - i);
    return res;
  }
  long long modexp(long long x, long long y) {
    long long res = 1;
    while (y > 0) {
      if (y & 1) res = (res * x) % mod;
      x = (x * x) % mod;
      y = y >> 1;
    }
    return res;
  }
  bool isOverflow(long long a, long long b) {
    if (a == 0 || b == 0) return false;
    long long result = a * b;
    if (a == result / b)
      return false;
    else
      return true;
  }
  void solve() {
    long long n, k, x;
    cin >> n >> k >> x;
    string s;
    cin >> s;
    vector<long long> v;
    for (long long i = 0; i < n; i++) {
      long long c = 0;
      while (i < n && s[i] == '*') i++, c++;
      if (c > 0) v.push_back(c), i--;
    }
    if (v.size() == 0) {
      cout << s << "\n";
      return;
    }
    long long curp = 0;
    for (long long i = 0; i < v.size(); i++) {
      v[i] = 1 + v[i] * k;
    }
    long long id = v.size() - 1;
    long long flag = 0;
    while (curp < x) {
      if ((double)v[id] > (double)INF / curp) {
        flag = 1;
        id--;
        break;
      }
      if (curp == 0)
        curp = v[id];
      else
        curp = curp * v[id];
      id--;
    }
    id++;
    if (flag == 0) curp = curp / v[id];
    vector<long long> b(v.size(), 0);
    long long y = x;
    for (long long i = id; i < v.size(); i++) {
      if (i == v.size() - 1) {
        b[i] = y;
        break;
      }
      b[i] = (y - 1) / (curp);
      y = y % curp;
      if (y == 0) y = curp;
      if (i + 1 < v.size()) curp = curp / v[i + 1];
    }
    b[b.size() - 1]--;
    string ans = "";
    long long ind = 0;
    for (long long i = 0; i < n; i++) {
      if (s[i] == 'a')
        ans.push_back('a');
      else {
        for (long long j = 0; j < b[ind]; j++) ans.push_back('b');
        ind++;
        while (i < n && s[i] == '*') i++;
        i--;
      }
    }
    cout << ans << "\n";
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long t = 1;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  