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.PrintStream;
import java.util.LinkedList;
import java.util.List;
import java.util.Scanner;

public class C {

    static long MAX = 1000000000000000000L;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        PrintStream out = System.out;

        int t = sc.nextInt();
        for (int i = 0; i < t; i++) {
            int n = sc.nextInt();
            int k = sc.nextInt();
            long x = sc.nextLong();
            String s = sc.next();
            out.println(solve(s, n, k, x));
        }

    }

    private static String solve(String s, int n, int k, long x) {
        StringBuilder builder = new StringBuilder();
        boolean wasStar = false;
        LinkedList<Integer> kList = new LinkedList<>();
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == 'a') {
                builder.append('a');
                kList.add(0);
                wasStar = false;
            } else {
                if (wasStar) {
                    kList.add(kList.removeLast() + k);
                } else {
                    builder.append('*');
                    kList.add(k);
                }
                wasStar = true;
            }
        }
        s = builder.toString();
        int[] ks = kList.stream().mapToInt(v -> v).toArray();

        n = ks.length;
        long[] d = new long[n + 1];
        d[n] = 1;
        for (int i = n - 1; i >= 0; i--) {
            if (s.charAt(i) == 'a') {
                d[i] = d[i + 1];
            } else {
                if (d[i + 1] == -1) {
                    d[i] = -1;
                    continue;
                }
                if ((ks[i] + 1) > MAX / d[i + 1]) {
                    d[i] = -1;
                } else {
                    d[i] = (ks[i] + 1) * d[i + 1];
                }
            }
        }


        StringBuilder res = new StringBuilder();
        for (int i = 0; i < n; i++) {
            if (s.charAt(i) == 'a') {
                res.append('a');
            } else {
                if (d[i + 1] != -1) {
                    long bCount = x / d[i + 1] - 1;
                    for (int j = 0; j < bCount; j++) {
                        res.append('b');
                    }
                    x -= bCount * d[i + 1];
                }
            }
        }

        return res.toString();
    }
}
