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.Closeable;
  import java.io.DataInputStream;
  import java.io.FileInputStream;
  import java.io.IOException;
  import java.util.Arrays;
  
  public class C {
      public static void main(String[] args) throws IOException {
          FastScanner scanner = new FastScanner();
          int t = scanner.nextInt();
          while (t-- > 0) {
              int n = scanner.nextInt(); int k = scanner.nextInt(); long x = scanner.nextLong();
              String s = scanner.next();
              int[] starCounts = new int[n];
              boolean prevA = true;
              int count = 0;
              for (char c : s.toCharArray()) {
                  if (c == '*') {
                      if (prevA) count++;
                      starCounts[count - 1]++;
                      prevA = false;
                  } else {
                      prevA = true;
                  }
              }
  
              if (count == 0) {
                  System.out.println(s);
                  continue;
              }
  
              x--;
  
              long[] bCount = new long[count];
              for (int i = count - 1; i > -1; i--) {
                  long v = starCounts[i] * (long)k + 1;
                  bCount[i] = x % v;
                  x /= v;
              }
  
              StringBuilder sb = new StringBuilder();
  
  
              int ptr = 0;
              prevA = true;
              for (char c : s.toCharArray()) {
                  if (c == '*') {
                      if (prevA) {
                          for (int i = 0; i < bCount[ptr]; i++) {
                              sb.append('b');
                          }
                          ptr++;
                      }
                      prevA = false;
                  } else {
                      sb.append('a');
                      prevA = true;
                  }
              }
  
              System.out.println(sb);
          }
      }
  
      /**
       * Credits: https://www.geeksforgeeks.org/fast-io-in-java-in-competitive-programming/
       */
      private static class FastScanner implements Closeable {
          final private int BUFFER_SIZE = 1 << 16;
          private DataInputStream din;
          private byte[] buffer;
          private int bufferPointer, bytesRead;
  
          public FastScanner() {
              din = new DataInputStream(System.in);
              buffer = new byte[BUFFER_SIZE];
              bufferPointer = bytesRead = 0;
          }
  
          public FastScanner(String file_name) throws IOException {
              din = new DataInputStream(new FileInputStream(file_name));
              buffer = new byte[BUFFER_SIZE];
              bufferPointer = bytesRead = 0;
          }
  
          public String readLine() throws IOException {
              byte[] buf = new byte[64];
              int cnt = 0, c;
              while ((c = read()) != -1) {
                  if (c == '\n' || c == '\r') {
                      if (cnt != 0) {
                          break;
                      } else {
                          continue;
                      }
                  }
                  buf[cnt++] = (byte) c;
              }
              return new String(buf, 0, cnt);
          }
  
          public String next() throws IOException {
              return next(32);
          }
  
          public String next(int len) throws IOException {
              byte[] buf = new byte[len];
              int ptr = 0;
              byte c = read();
              while (c <= ' ') c = read();
              do {
                  if (ptr == buf.length) {
                      buf = Arrays.copyOf(buf, buf.length << 1);
                  }
                  buf[ptr++] = c;
              } while ((c = read()) > ' ');
              if (buf.length == ptr) return new String(buf);
              return new String(Arrays.copyOf(buf, ptr));
          }
  
          public int nextInt() throws IOException {
              int ret = 0;
              byte c = read();
              while (c <= ' ') c = read();
              boolean neg = (c == '-');
              if (neg) c = read();
              do {
                  ret = ret * 10 + c - '0';
              } while ((c = read()) >= '0' && c <= '9');
              if (neg) return -ret;
              return ret;
          }
  
          public long nextLong() throws IOException {
              long ret = 0;
              byte c = read();
              while (c <= ' ') c = read();
              boolean neg = (c == '-');
              if (neg) c = read();
              do {
                  ret = ret * 10 + c - '0';
              } while ((c = read()) >= '0' && c <= '9');
              if (neg) return -ret;
              return ret;
          }
  
          public double nextDouble() throws IOException {
              double ret = 0, div = 1;
              byte c = read();
              while (c <= ' ') c = read();
              boolean neg = (c == '-');
              if (neg) c = read();
              do {
                  ret = ret * 10 + c - '0';
              } while ((c = read()) >= '0' && c <= '9');
              if (c == '.') {
                  while ((c = read()) >= '0' && c <= '9') {
                      ret += (c - '0') / (div *= 10);
                  }
              }
              if (neg) return -ret;
              return ret;
          }
  
          private void fillBuffer() throws IOException {
              bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
              if (bytesRead == -1) buffer[0] = -1;
          }
  
          private byte read() throws IOException {
              if (bufferPointer == bytesRead) fillBuffer();
              return buffer[bufferPointer++];
          }
  
          @Override
          public void close() throws IOException {
              if (din == null) return;
              din.close();
          }
      }
  }
  