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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class CF_1620C {
	public static void main(String[] args) throws NumberFormatException, IOException {
		MyReader reader = new MyReader();
		int t = reader.nextInt();
		for(int testcase=0; testcase<t; testcase++) {
			int n = reader.nextInt();
			int k = reader.nextInt();
			long x = reader.nextLong()-1;
			String s = reader.next();
			ArrayList<Integer> consecAsterisks = new ArrayList<>();
			int consec = 0;
			for(int i=0; i<n; i++) {
				if(s.charAt(i) == '*') consec++;
				else {
					consecAsterisks.add(consec);
					consec = 0;
				}
			}
			consecAsterisks.add(consec);
			
			StringBuilder sb = new StringBuilder();
			long rank = 0;
			for(int i=0; i<consecAsterisks.size(); i++) {
				//skip: to avoid overflow, if ways is too large, we won't be able to progress
				//this is sloppy, so we still need to be careful later. This is just for overflow
				boolean skip = false;
				long ways = 1;
				for(int j=i+1; j<consecAsterisks.size(); j++) {
					if(2_000_000_000_000_000_000L/ways+2 < consecAsterisks.get(j)*k+1) {
						skip = true;
					}
					else {
						ways *= consecAsterisks.get(j)*k+1;
					}
				}
				
				if(skip) {
					sb.append("a");
					continue;
				}
				
				
				long numb = (x-rank)/ways;
				for(int j=0; j<numb; j++) {
					sb.append("b");
				}
				rank += numb * ways;
				if(i != consecAsterisks.size()-1) sb.append("a");
			}
			System.out.println(sb.toString());
		}
	}
	
	static class MyReader {
        BufferedReader br;
        StringTokenizer st;
        public MyReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        public String next() throws IOException  {
            while(st == null || !st.hasMoreElements()) {
                st = new StringTokenizer(br.readLine());
            }
            return st.nextToken();
        }
        public int nextInt() throws NumberFormatException, IOException {
            return Integer.parseInt(next());
        }
        public long nextLong() throws NumberFormatException, IOException {
            return Long.parseLong(next());
        }
    }
}
