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: import java.io.*;
  import java.util.*;
  
  public class Main {
      class JoinSet {
          int[] fa;
          JoinSet(int n){
              fa = new int[n];
              for(int i = 0;i < n;i++) fa[i] = i;
          }
          int find(int x){
              if(x != fa[x]) fa[x] = find(fa[x]);
              return fa[x];
          }
          void join(int x, int y){
              x = find(x);
              y = find(y);
              fa[x] = y;
          }
      }
      static BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
      static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
      static int mod = (int)1e9+7;
      static boolean[] isprime=new boolean[1];
      static {
          for(int i=2;i<isprime.length;i++) isprime[i]=true;
          for(int i=2;i<isprime.length;i++) {
              if(isprime[i]) {
                  for(int k = 2; i*k<isprime.length; k++) {
                      isprime[i*k] = false;
                  }
              }
          }
      }
      public static long gcd(long a, long b){
          return b == 0 ? a : gcd(b,a%b);
      }
      public static long lcm(long a, long b){
          return a*b / gcd(a,b);
      }
      public static int get() throws Exception{
          String ss = bf.readLine();
          if (ss.contains(" ")) ss = ss.substring(0, ss.indexOf(" "));
          return Integer.parseInt(ss);
      }
      public static int[] getInt() throws Exception {
          String s[] = bf.readLine().split(" ");
          int[] a = new int[s.length];
          for(int i = 0;i < a.length;i++){
              a[i] = Integer.parseInt(s[i]);
          }
          return a;
      }
      public static long[] getLong() throws Exception {
          String s[] = bf.readLine().split(" ");
          long[] a = new long[s.length];
          for(int i = 0;i < a.length;i++){
              a[i] = Long.parseLong(s[i]);
          }
          return a;
      }
      public static void print(int[] a) throws Exception {
          for(int i : a){
              bw.write(i + " ");
          }
          println();
      }
      public static void print(long[] a) throws Exception {
          for(long i : a){
              bw.write(i + " ");
          }
          println();
      }
      public static void println() throws Exception{
          bw.write("\n");
      }
      public static long pow(long a, long b){
          long ans = 1;
          while(b > 0){
              if((b&1) == 1){
                  ans *= a;
              }
              a*=a;
              b>>=1;
          }
          return ans;
      }
      public static int powmod(long a, long b, int mod){
          long ans = 1;
          while(b > 0){
              if((b&1) == 1){
                  ans = ans * a % mod;
              }
              a = a * a % mod;
              b>>=1;
          }
          return (int)ans;
      }
      public static void sort(int[] a){
          int n = a.length;
          Integer[] b = new Integer[n];
          for(int i = 0;i < n;i++) b[i] = a[i];
          Arrays.sort(b);
          for(int i = 0;i < n;i++) a[i] = b[i];
      }
  
      public static void main(String[] args) throws Exception {
          int t = get();
          while (t-- > 0){
              long[] a = getLong();
              long k = a[1], x = a[2];
              x--;
              StringBuilder ans = new StringBuilder();
              String s = bf.readLine();int n = s.length();
              for(int i = n-1;i >= 0;i--){
                  if(s.charAt(i) == '*'){
                      int cnt = 0;
                      while(i >= 0 && s.charAt(i) == '*'){
                          cnt++;
                          i--;
                      }
                      i++;
                      long d = cnt*k+1;
                      long y = x%d;
                      for(int j = 0;j < y;j++) ans.append('b');
                      x /= d;
                  }else ans.append('a');
              }
              bw.write(ans.reverse().toString() + "\n");
          }
          bw.flush();
      }
  }
  