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.*;
import java.math.*;

public class Codeforces
{
    public static void main(String args[])throws Exception
    {
        BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb=new StringBuilder();
        int t=Integer.parseInt(bu.readLine());
        while(t-->0)
        {
            String s[]=bu.readLine().split(" ");
            int n=Integer.parseInt(s[0]),k=Integer.parseInt(s[1]); long x=Long.parseLong(s[2]);
            char c[]=bu.readLine().toCharArray();
            if(x==0)
            {
                for(int i=0;i<n;i++)
                if(c[i]=='a') sb.append('a');
                sb.append("\n");
                continue;
            }

            int i,l=0;
            ArrayList<Integer> ci=new ArrayList<>();
            for(i=0;i<n;i++)
            if(c[i]=='*') l++;
            else
            {
                if(l!=0) ci.add(l*k+1);
                l=0;
            }
            if(l!=0) ci.add(l*k+1);

            int sz=ci.size();
            long ri[]=new long[sz+1];
            BigInteger mul=new BigInteger("1");
            for(i=sz-1;i>=0;i--)
            {
                mul=mul.multiply(mul.valueOf(ci.get(i)));
                //System.out.println(mul+" "+mul.compareTo(mul.valueOf(x)));
                if(mul.compareTo(mul.valueOf(x))>0) break;
                ri[i]=mul.longValue();
            }
            while(i>=0) ri[i--]=(long)2e18;
            ri[sz]=1;

            int add[]=new int[sz+1];
            x--;
            while(x>0)
            {
                int r=rightmost(ri,sz,x);
                add[r-1]++;
                x-=ri[r];
            }

            StringBuilder sb2=new StringBuilder();
            int j=sz-1;
            for(i=n-1;i>=0;i--)
            {
                int y=i;
                while(y>=0 && c[y]=='*') y--;
                if(y!=i)
                {
                    while(add[j]-->0) sb2.append('b');
                    i=y+1;
                    j--;
                }
                else sb2.append('a');
            }
            sb.append(sb2.reverse()+"\n");
        }
        System.out.print(sb);
    }

    static int rightmost(long a[],int r,long x)
    {
        int l=0,mid,ans=r;
        while(l<=r)
        {
            mid=(l+r)>>1;
            if(a[mid]>x)
            {
                ans=mid+1;
                l=mid+1;
            }
            else r=mid-1;
        }
        return ans;
    }
}