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 int maxn = 2e5 + 7, inf = 2e9 + 9, mod = 1e9 + 7;
  const long long linf = 2e18 + 7;
  int n, k;
  long long x, d[maxn];
  void solve() {
    vector<long long> a;
    char s[maxn], dum[maxn];
    scanf("%d%d%I64d", &n, &k, &x);
    memset(d, 0, (n + 3) * sizeof(long long));
    scanf("%s", &dum);
    int l = 0;
    for (int i = 0; i < n + 1; ++i) {
      if (dum[i] == '*')
        l++;
      else {
        a.push_back(1ll * l * k);
        l = 0;
      }
    }
    int sz = a.size();
    a.push_back(0);
    long long t = 1;
    for (int i = sz - 1; i >= 0; --i) {
      if (linf / t < (a[i] + 1) or t * (a[i] + 1) >= x) {
        x--;
        for (int j = i; j < sz; ++j) {
          d[j] = x / t;
          x -= d[j] * t;
          t /= a[j + 1] + 1;
        }
        break;
      }
      t = t * (a[i] + 1);
    }
    for (int i = 0; i < sz; ++i) {
      for (int j = 0; j < d[i]; ++j) {
        printf("b");
      }
      if (i != sz - 1) printf("a");
    }
    puts("");
  }
  int main() {
    int test;
    scanf("%d", &test);
    while (test--) solve();
    return 0;
  }
  