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 mod = 1000000007;
  int ioo = 1000000000;
  long long loo = 1000000000000000000;
  void add_self(int &x, int y) {
    if ((x += y) >= mod) x -= mod;
  }
  int add(int x, int y) { return add_self(x, y), x; }
  void sub_self(int &x, int y) {
    if ((x -= y) < 0) x += mod;
  }
  int sub(int x, int y) { return sub_self(x, y), x; }
  int mul(int x, int y) { return (long long)x * y % mod; }
  long long po(long long k, long long n) {
    long long res = 1;
    while (n) {
      if (n % 2) res *= k;
      k *= k;
      n /= 2;
    }
    return res;
  }
  long long pom(long long k, long long n) {
    long long res = 1;
    while (n) {
      if (n % 2) res = mul(res, k);
      k = mul(k, k);
      n /= 2;
    }
    return res;
  }
  char a[2005];
  vector<long long> v;
  long long n, k, x;
  int main() {
    int tt;
    scanf("%d", &tt);
    while (tt--) {
      v.clear();
      scanf("%lld%lld%lld", &n, &k, &x);
      scanf("%s", a);
      long long cnt = 0;
      for (int i = 0; i < n; i++) {
        if (a[i] == 'a') {
          if (cnt) v.push_back(cnt * k);
          cnt = 0;
        } else
          cnt++;
      }
      if (cnt) v.push_back(cnt * k);
      vector<long long> b = v;
      if (!b.empty()) b.back()++;
      int idx = b.size();
      for (int i = (int)b.size() - 2; i >= 1; i--) {
        if (b[i] + 1 <= (loo * 2) / b[i + 1])
          b[i] = (b[i] + 1) * b[i + 1];
        else {
          idx = i;
          break;
        }
      }
      if (idx != b.size())
        for (int i = idx; i >= 0; i--) b[i] = loo * 2;
      deque<long long> ans(v.size(), 0);
      for (int i = 0; i < v.size(); i++) {
        if (i == (int)v.size() - 1)
          ans[i] = x - 1;
        else {
          long long steps = min((x - 1) / b[i + 1], v[i]);
          ans[i] = steps;
          x -= steps * b[i + 1];
        }
      }
      for (int i = 0; i < n; i++) {
        if (a[i] == 'a')
          printf("a");
        else {
          if (i == 0 || a[i] != a[i - 1]) {
            long long cur = ans.front();
            while (cur--) printf("b");
            ans.pop_front();
          }
        }
      }
      printf("\n");
    }
    return 0;
  }
  