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: //package Div2.C;
  
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  
  public class BAString {
      public static void main(String[] args) throws IOException {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          int t = Integer.parseInt(br.readLine());
          while (t > 0) {
              String[] nkx = br.readLine().split(" ");
              int n = Integer.parseInt(nkx[0]);
              int k = Integer.parseInt(nkx[1]);
              long x = Long.parseLong(nkx[2]);
              String s = br.readLine();
  
              if (x == 1) {
                  StringBuilder ans = new StringBuilder();
                  for (int i = 0; i < n; i++) {
                      if (s.charAt(i) != '*')
                          ans.append(s.charAt(i));
                  }
                  System.out.println(ans);
              } else {
                  //there should be some asterisks
                  long abStrings = 1;
                  int nextIterationStartPos = -1;
                  for (int i = n - 1; i >= 0; i--) {
                      if (s.charAt(i) == '*') {
                          int cnt = 1;
                          while (i - 1 >= 0 && s.charAt(i - 1) == '*') {
                              cnt += 1;
                              i--;
                          }
  //                        if (abStrings == 0)
  //                            abStrings = cnt * k + 1;
  //                        else {
  //                            long temp = abStrings * (cnt * k + 1);
  //                            assert(temp>=abStrings);
  //                            abStrings = temp;
  //                        }
                          long temp = abStrings * (cnt * k + 1);
                          if (temp < 0 || temp < abStrings) {
                              nextIterationStartPos = i;
                              break;
                          } else if (x <= temp) {
                              nextIterationStartPos = i;
                              break;
                          }
                          abStrings = temp;
                      }
                  }
                  StringBuilder ans = new StringBuilder();
                  for (int i = 0; i < nextIterationStartPos; i++) {
                      if (s.charAt(i) != '*')
                          ans.append(s.charAt(i));
                  }
                  while (nextIterationStartPos + 1 < n && s.charAt(nextIterationStartPos + 1) == '*') {
                      nextIterationStartPos += 1;
                  }
                  nextIterationStartPos++;
                  long q = x / abStrings;
                  long rem = x % abStrings;
                  if (rem > 0) {
                      q = q + 1;
                      x = rem;
                  } else {
                      x = abStrings;
                  }
                  while (q > 1) {
                      ans.append('b');
                      q--;
                  }
  
                  for (int i = nextIterationStartPos; i < n; i++) {
                      if (s.charAt(i) == '*') {
                          int cnt = 1;
                          while (i + 1 < n && s.charAt(i + 1) == '*') {
                              cnt += 1;
                              i++;
                          }
                          abStrings = abStrings / (cnt * k + 1);
                          q = x / abStrings;
                          rem = x % abStrings;
                          if (rem > 0) {
                              q = q + 1;
                              x = rem;
                          } else {
                              x = abStrings;
                          }
                          while (q > 1) {
                              ans.append('b');
                              q--;
                          }
                      } else
                          ans.append(s.charAt(i));
                  }
  
                  System.out.println(ans);
              }
  
              t--;
          }
      }
  }
  