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: /*
   * Everything is Hard 
   * Before Easy 
   * Jai Mata Dii 
   */ 
   
  import java.util.*;
  import java.io.*; 
    
  public class Main {
  	static class FastReader{ BufferedReader br;StringTokenizer st;public FastReader(){br = new BufferedReader(new InputStreamReader(System.in));}String next(){while (st == null || !st.hasMoreElements()){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; }} 
  	static long mod = (long)(1e9+7); 
  //	 static long mod = 998244353; 
  //	 static Scanner sc = new Scanner(System.in); 
  	static FastReader sc = new FastReader(); 
  	static PrintWriter out = new PrintWriter(System.out);
  	public static void main (String[] args) {
  		int ttt = 1;
  		ttt = sc.nextInt();
  		z :for(int tc=1;tc<=ttt;tc++){
  			int n = sc.nextInt();
  			int k = sc.nextInt();
  			long x = sc.nextLong();
  			
  			x-=1;
  			
  			char s[] = sc.next().toCharArray();
  
  			if(x == 0) {
  				for(char c : s) {
  					if(c == 'a')
  						out.write("a");
  				}
  				out.write("\n");
  				continue;
  			}
  			
  			ArrayList<Long> a = new ArrayList<>();
  			
  			for(int i=0;i<s.length;i++) {
  				if(s[i] == 'a') {
  					a.add(-1L);
  				}
  				else {
  					 if(a.size()==0 || a.get(a.size()-1)==-1) {
  						 a.add((long) (k+1));
  					 }
  					 else {
  						 a.set(a.size()-1, a.get(a.size()-1)+k);
  					 }
  				}
  			}
  			
  			long add[] = new long[a.size()];
  			long prev = 1;
  			
  			for(int i=a.size()-1;i>=0;i--) {
  				if(a.get(i) == -1) {
  					add[i] = -1;
  					continue;
  				}
  				long cur = a.get(i);
  				add[i] = x%cur;
  				x = x/cur;
  			}
  			
  			StringBuilder form = new StringBuilder();
  			
  			for(int i=0;i<add.length;i++) {
  				if(add[i] == -1) form.append("a");
  				else {
  					while(add[i]-->0) {
  						form.append("b");
  					}
  				}
  			}
  			out.write(form+"\n");
  		}
  		out.close();
  	}
  	static long pow(long a, long b){long ret = 1;while(b>0){if(b%2 == 0){a = (a*a)%mod;b /= 2;}else{ret = (ret*a)%mod;b--;}}return ret%mod;}
  	static long gcd(long a,long b){if(b==0) return  a; return gcd(b,a%b); } 
  	private static void sort(int[] a) {List<Integer> k = new ArrayList<>();for(int val : a) k.add(val);Collections.sort(k);for(int i=0;i<a.length;i++) a[i] = k.get(i);} 
  	private static void ini(List<Integer>[] tre2){for(int i=0;i<tre2.length;i++){tre2[i] = new ArrayList<>();}} 
  	private static void init(List<int[]>[] tre2){for(int i=0;i<tre2.length;i++){tre2[i] = new ArrayList<>();}} 
  	private static void sort(long[] a) {List<Long> k = new ArrayList<>();for(long val : a) k.add(val);Collections.sort(k);for(int i=0;i<a.length;i++) a[i] = k.get(i);} 
  }