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.
2
55 982 31372
a*a*aa*aa*a*aa*aa*aaa*aaaaa*a*aa*aaaaa*aaaaa*aa*a*a*a*a
63 9 10946
*aaaaa*aa*a*a*aa*aaaa*a*aa*a*a*aaaa*aaaaa*a*a*aaa*aa*aaa*a*a*aa

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbba
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabaaaaabbbbbbbbbabbbbabbbbbaa

import java.io.*;
import java.util.*;

public class Main {
	
	public static void main(String[] args) throws Exception {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));		
		BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));		
		
		int T = Integer.parseInt(in.readLine());
		char[] ss = new char[2020];
		int[] count = new int[2020];
		
		long[] dp = new long[2020]; //may overflow
		
		for (int ts=1; ts<=T; ts++) {
			
			String[] line = in.readLine().split(" ");
			int N = Integer.parseInt(line[0]);
			int K = Integer.parseInt(line[1]);
			long X = Long.parseLong(line[2]);
			
			String S = in.readLine();
			int size = 0;

			{
				int idx = 0;
				while (idx < S.length()) {
					if (S.charAt(idx) == 'a') {
						ss[size] = 'a';
						count[size] = 1;
						size++;
						idx++;
						continue;
					}
					
					int left = idx;
					while (idx < S.length() && S.charAt(idx) == '*') {
						idx++;
					}
					ss[size] = '*';
					count[size] = idx-left;
					size++;
				}
			}
			
			
			dp[size] = 1;
			for (int i=size-1; i>=0; i--) {
				if (ss[i] == 'a') 
					dp[i] = dp[i+1];
				else {
					if (1.0 * dp[i+1] > (1.0 * X / (count[i] * K + 1) + 1e-18)) {
						dp[i] = X+1;
					} else {
						dp[i] = dp[i+1] * (count[i] * K + 1);
					}
				}
			}
			
			StringBuilder result = new StringBuilder();
			for (int i=0; i<size; i++) {
				if (ss[i] == 'a') {
					result.append('a');
					continue;
				}
				
				for (int cnt=0; cnt<=(count[i] * K); cnt++) {
					if (X > dp[i+1]) {
						X-=dp[i+1];
						result.append('b');
					} else {
						break;
					}
				}
			}
			
			out.write(result.toString() + "\n");
			out.flush();
		}
		
		in.close();
		out.close();
	}

}
