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 codeforces.edu119;
  
  import java.io.*;
  import java.util.*;
  
  import static java.lang.Math.*;
  
  //Think through the entire logic before jump into coding!
  //If you are out of ideas, take a guess! It is better than doing nothing!
  //Read both C and D, it is possible that D is easier than C for you!
  
  //Be aware of integer overflow!
  //If you find an answer and want to return immediately, don't forget to flush before return!
  
  public class C {
      static InputReader in;
      static PrintWriter out;
  
      public static void main(String[] args) {
          //initReaderPrinter(true);
          initReaderPrinter(false);
          solve(in.nextInt());
          //solve(1);        
      }
  
      static void solve(int testCnt) {
          for (int testNumber = 0; testNumber < testCnt; testNumber++) {
              int n = in.nextInt(), k = in.nextInt();
              long x = in.nextLong() - 1;
              char[] s = in.next().toCharArray();
              int bCnt = 0;
              ArrayDeque<int[]> base = new ArrayDeque<>();
              for(int i = 0; i < n; i++) {
                  if(s[i] == 'a') {
                      if(bCnt > 0) {
                          base.addLast(new int[]{i - bCnt, bCnt * k + 1});
                      }
                      bCnt = 0;
                  }
                  else {
                      bCnt++;
                  }
              }
              if(bCnt > 0) {
                  base.addLast(new int[]{n - bCnt, bCnt * k + 1});
              }
              ArrayDeque<int[]> bq = new ArrayDeque<>();
              while(x > 0) {
                  int[] e = base.removeLast();
                  bq.addFirst(new int[]{e[0], (int)(x % e[1])});
                  x /= e[1];
              }
              StringBuilder ans = new StringBuilder();
              for(int i = 0; i < n; i++) {
                  if(s[i] == 'a') ans.append('a');
                  else if((i == 0 || s[i] != s[i - 1]) && bq.size() > 0 && bq.peekFirst()[0] == i) {
                      int total = bq.removeFirst()[1];
                      for(int j = 0; j < total; j++) {
                          ans.append('b');
                      }
                  }
              }
              out.println(ans);
          }
          out.close();
      }
  
      static void initReaderPrinter(boolean test) {
          if (test) {
              try {
                  in = new InputReader(new FileInputStream("src/input.in"));
                  out = new PrintWriter(new FileOutputStream("src/output.out"));
              } catch (IOException e) {
                  e.printStackTrace();
              }
          } else {
              in = new InputReader(System.in);
              out = new PrintWriter(System.out);
          }
      }
  
      static class InputReader {
          BufferedReader br;
          StringTokenizer st;
  
          InputReader(InputStream stream) {
              try {
                  br = new BufferedReader(new InputStreamReader(stream), 32768);
              } catch (Exception e) {
                  e.printStackTrace();
              }
          }
  
          String next() {
              while (st == null || !st.hasMoreTokens()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
  
          double nextDouble() {
              return Double.parseDouble(next());
          }
  
          String nextLine() {
              String str = "";
              try {
                  str = br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
  
          Integer[] nextIntArray(int n) {
              Integer[] a = new Integer[n];
              for (int i = 0; i < n; i++) a[i] = nextInt();
              return a;
          }
  
          int[] nextIntArrayPrimitive(int n) {
              int[] a = new int[n];
              for (int i = 0; i < n; i++) a[i] = nextInt();
              return a;
          }
  
          int[] nextIntArrayPrimitiveOneIndexed(int n) {
              int[] a = new int[n + 1];
              for (int i = 1; i <= n; i++) a[i] = nextInt();
              return a;
          }
  
          Long[] nextLongArray(int n) {
              Long[] a = new Long[n];
              for (int i = 0; i < n; i++) a[i] = nextLong();
              return a;
          }
  
          long[] nextLongArrayPrimitive(int n) {
              long[] a = new long[n];
              for (int i = 0; i < n; i++) a[i] = nextLong();
              return a;
          }
  
          long[] nextLongArrayPrimitiveOneIndexed(int n) {
              long[] a = new long[n + 1];
              for (int i = 1; i <= n; i++) a[i] = nextLong();
              return a;
          }
  
          String[] nextStringArray(int n) {
              String[] g = new String[n];
              for (int i = 0; i < n; i++) g[i] = next();
              return g;
          }
  
          List<Integer>[] readGraphOneIndexed(int n, int m) {
              List<Integer>[] adj = new List[n + 1];
              for (int i = 0; i <= n; i++) {
                  adj[i] = new ArrayList<>();
              }
              for (int i = 0; i < m; i++) {
                  int u = nextInt();
                  int v = nextInt();
                  adj[u].add(v);
                  adj[v].add(u);
              }
              return adj;
          }
  
          List<Integer>[] readGraphZeroIndexed(int n, int m) {
              List<Integer>[] adj = new List[n];
              for (int i = 0; i < n; i++) {
                  adj[i] = new ArrayList<>();
              }
              for (int i = 0; i < m; i++) {
                  int u = nextInt() - 1;
                  int v = nextInt() - 1;
                  adj[u].add(v);
                  adj[v].add(u);
              }
              return adj;
          }
  
          /*
              A more efficient way of building a graph using int[] instead of ArrayList to store each node's neighboring nodes.
              1-indexed.
           */
          int[][] buildGraph(int nodeCnt, int edgeCnt) {
              int[] end1 = new int[edgeCnt], end2 = new int[edgeCnt];
              int[] edgeCntForEachNode = new int[nodeCnt + 1], idxForEachNode = new int[nodeCnt + 1];
              for (int i = 0; i < edgeCnt; i++) {
                  int u = in.nextInt(), v = in.nextInt();
                  edgeCntForEachNode[u]++;
                  edgeCntForEachNode[v]++;
                  end1[i] = u;
                  end2[i] = v;
              }
              int[][] adj = new int[nodeCnt + 1][];
              for (int i = 1; i <= nodeCnt; i++) {
                  adj[i] = new int[edgeCntForEachNode[i]];
              }
              for (int i = 0; i < edgeCnt; i++) {
                  adj[end1[i]][idxForEachNode[end1[i]]] = end2[i];
                  idxForEachNode[end1[i]]++;
                  adj[end2[i]][idxForEachNode[end2[i]]] = end1[i];
                  idxForEachNode[end2[i]]++;
              }
              return adj;
          }
      }
  }