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

#include<bits/stdc++.h>
using namespace std;
#define int long long

signed main()
{
    ios::sync_with_stdio(false) ;
    cin.tie(0) ;
    cout.tie(0) ;

    int t;
    cin>>t;
    while(t--)
    {
        int n,k,x;
        cin>>n>>k>>x;
        //x--;
        string s;
        cin>>s;
        int pstarcount=0;
        int starcount=0;
        int val=1;
        //v.push_back(0);
        vector<int> v;
        for(int i=n-1;i>=0;i--)
        {
            if(s[i]=='*')
            {
                pstarcount++;
            }
            else
            {   
                if(pstarcount)
                {
                    starcount++;
                    val*=(pstarcount*k+1);
                    v.push_back(val);
                    pstarcount=0;
                }
                
            }
        }
        if(pstarcount)
        {
            starcount++;
            val*=(pstarcount*k+1);
            v.push_back(val);
        }
        int pos=starcount;
        int count[starcount]={0};
        pos--;
        int end=v.size();
        end--;
        for(int i=0;i<=end;i++)
            cout<<v[i]<<' ';
        cout<<'\n';
        while(x)
        {
            int hi=end;
            int lo=0;
            int ans=-1;
            while(lo<=hi)
            {
                int mid=(hi+lo)/2;
                if(x>=v[mid])
                {
                    ans=mid;
                    lo=mid+1;
                }
                else
                {
                    hi=mid-1;
                }
            }
            if(ans==-1)
            {
                count[pos]=x-1;
                x=0;
            }
            else 
            {   
                //cout<<ans<<' ';
                if(x==v[ans])
                {
                    if(ans)
                        count[pos-ans]=(v[ans]/v[ans-1]-1);
                    else count[pos-ans]=(v[ans]-1);
                    x=0;
                }
                else
                {
                    //cout<<x<<' '<<v[ans]<<'\n';
                    count[pos-ans-1]=(x-1)/v[ans];
                    x-=v[ans]*((x-1)/v[ans]);
                }   
            }
        }
        //cout<<'\n';
        // for(int i=0;i<starcount;i++)
        //     cout<<count[i]<<' ';
        // cout<<'\n';
        bool prev=false;
        int posi=0;
        for(int i=0;i<n;i++)
        {
            if(s[i]=='a')
            {
                if(prev)
                {
                    for(int j=0;j<count[posi];j++)
                        cout<<'b';
                    posi++;
                }
                cout<<'a';
                prev=false;
            }
            else 
            {
                prev=true;
            }
        }
        if(prev)
        {
            for(int j=0;j<count[posi];j++)
                    cout<<'b';
            posi++;
        }
        cout<<'\n';
    }

    return 0;
}