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;
  using i64 = long long;
  using i32 = int;
  const int INF = 0x3f3f3f3f;
  const double tol = 1e-8;
  const int P = 998244353;
  int pop_count(i64 x) { return __builtin_popcountll(x); }
  int pop_count(int x) { return __builtin_popcount(x); }
  template <typename S, typename T>
  bool check_and_save_min_in_lhs(S& lhs, const T& rhs) {
    return lhs < rhs ? 1 : lhs = rhs, 0;
  }
  template <typename S, typename T>
  bool check_and_save_max_in_lhs(S& lhs, const T& rhs) {
    return lhs > rhs ? 1 : lhs = rhs, 0;
  }
  template <typename T>
  T min_all(T ele) {
    return ele;
  }
  template <typename T, typename... Args>
  T min_all(T fst_el, Args... others) {
    return min_all(fst_el, mins(others...));
  }
  template <typename T>
  T maxs(T ele) {
    return ele;
  }
  template <typename T, typename... Args>
  T maxs(T fst_el, Args... others) {
    return max(fst_el, maxs(others...));
  }
  template <typename T>
  void read_int(T& num) {
    bool positive = true;
    char ch;
    num = 0;
    while (isspace(ch = getchar()))
      ;
    if (ch == '-')
      positive = false;
    else
      num = ch - '0';
    while ((ch = getchar()) >= '0' && ch <= '9') num = num * 10 + ch - '0';
    if (!positive) num = -num;
  }
  void read_ints(){};
  template <typename T, typename... Args>
  void read_ints(T& fst_el, Args&... others) {
    read_int(fst_el);
    read_ints(others...);
  }
  template <typename T>
  void int_out(T num) {
    if (num < 0) {
      putchar('-');
      num = -num;
    }
    if (num > 9) int_out(num / 10);
    putchar(num % 10 + '0');
  }
  void int_outs() { puts(""); }
  template <typename T, typename... Args>
  void int_outs(T fst_ele, Args... others) {
    int_out(fst_ele);
    putchar(' ');
    int_outs(others...);
  }
  const int N = 2002;
  char s[N];
  int main() {
    int t;
    read_int(t);
    while (t--) {
      int n, k;
      i64 x;
      read_ints(n, k);
      read_int(x);
      gets(s);
      int len = 0, cur = 0;
      char ch[3] = "a*";
      vector<int> r[2];
      for (int i = 0, len; i < n; i += len) {
        len = 0;
        while (i + len < n && s[i + len] == ch[cur]) {
          ++len;
        }
        r[cur].push_back(len);
        cur ^= 1;
      }
      if (r[0].size() != r[1].size()) r[1].push_back(0);
      vector<i64> acc(r[1].size(), 0);
      int idx;
      i64 cnt = 1;
      for (idx = r[1].size() - 1; idx >= 0; --idx) {
        acc[idx] = cnt;
        if ((i64(k) * r[1][idx] + 1) > x / cnt) {
          break;
        } else {
          cnt = cnt * (1 + (i64)k * r[1][idx]);
        }
      }
      for (int i = 0; i < r[0].size(); ++i) {
        while (r[0][i]--) {
          putchar('a');
        }
        if (i >= idx && x > 0) {
          cnt = (x - 1) / acc[i];
          x -= cnt * acc[i];
          for (int j = 0; j < cnt; ++j) {
            putchar('b');
          }
        }
      }
      puts("");
    }
    return 0;
  }
  