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

public class C {
    static MyScanner sc;
    static PrintWriter out;
    static {
        sc = new MyScanner();
        out = new PrintWriter(System.out);
    }
    public static ArrayList<Pair> run_encode(char[] a) {
        int n = a.length;
        ArrayList<Pair> ans = new ArrayList<>();
        char prev = a[0];
        int c = 1;
        for(int i = 1; i < n; i++) {
            if(a[i] == prev) c++;
            else{
                ans.add(new Pair(prev, c));
                c = 1;
                prev = a[i];
            }
        }
        ans.add(new Pair(prev, c));
        return ans;
    }
    public static void solve() {
        int n = sc.nextInt();
        int k = sc.nextInt();
        long x = sc.nextLong() - 1;
        char[] a = sc.next().toCharArray();
        StringBuilder ans = new StringBuilder();
        ArrayList<Pair> A = run_encode(a);
        for(int i = A.size() - 1; i >= 0; i--) {
            Pair p = A.get(i);
            if(p.x == 'a')
                p.res = p.y;
            else {
                p.y = p.y * k + 1;
                p.res = (int)x % p.y;
                x = x / p.y;
            }
        }
        for(Pair p : A) {
            for(int i = 0; i < p.res; i++)
                out.print((p.x == 'a' ? 'a' : 'b'));
        }
        out.println();
    }
    public static void main(String[] args) {
        int t = sc.nextInt();
        while(t-- > 0)
            solve();
        out.flush();
    }
}

class Pair {
    char x;
    int y, res;
    Pair(char a, int b) {
        x = a;
        y = b;
        res = 0;
    }
    public String toString() {
        return "(" + x +", " + y + ", " + res + ")";
    }
}

class MyScanner {
    BufferedReader br;
    StringTokenizer tok;
    MyScanner() {
        try { br = new BufferedReader(new InputStreamReader(System.in)); }
        catch(Exception e) { System.out.println(e); }
        tok = new StringTokenizer("");
    }
    public String next() {
        try {
            while(!tok.hasMoreTokens()) tok = new StringTokenizer(br.readLine());
        }
        catch(Exception e) { System.out.println(e); }
        return tok.nextToken();
    }
    public int nextInt() {
        return Integer.parseInt(next());
    }
    public long nextLong() {
        return Long.parseLong(next());
    }
    public double nextDouble() {
        return Double.parseDouble(next());
    }
}
