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;

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

#define take(i)                    int i;cin>>i
#define taken                    take(n)
#define gcd(a,b)                __gcd(a,b)
#define endl                    '\n'
#define f                        first
#define s                        second
#define pb                        push_back
#define rep(i,a,b,c)            for(int i=a;i<b;i+=c)
#define ub                        upper_bound
#define lb                        lower_bound
#define bs                        binary_search
#define all(x)                    x.begin(), x.end()
#define mp                        make_pair
#define PI                        3.141592653589793238
#define E                        2.7182818284590452353602874713527
#define pi                        pair<int,int>
#define vi                        vector<int>
typedef unsigned long int    uli;
typedef long long        ll;
typedef unsigned long long    ull;
const unsigned int M = 1000000007;

using namespace std;

vi primes;

ull binpow(ll a,ll b,ll m);
void primesfill(int k);

void solve(){
    ll n,k,x;
    cin>>n>>k>>x;
    ll j=1;
    string s;
    cin>>s;
    string ans="";
    // for(int i=n-1;i>=0;i--){
    //     if(s[i]=='a'){ans="a"+ans;continue;}
    //     if(j*k<x){
    //         j*=k;
    //         rep(l,0,k,1)ans="b"+ans;
    //     }
    //     else{
    //         int t=1;
    //         while(j*t<x){
    //             t++;
    //             ans="b"+ans;
    //         }
    //         j=j*t;
    //     }
    // }
    ll ct=0;
    for(int i=n-1;i>=0;i--){
        if(s[i]=='*'){ct++;continue;}
        if(j+j*k*ct>x){
            ll t=1;
            while(j*t<x){
                t++;
                ans="b"+ans;
            }
            j=j*t;
        }
        else if(ct){
            j*=(ct*k+1);
            for(int i=0;i<ct*k;i++)ans="b"+ans;
        }
        ans="a"+ans;
        ct=0;
    }
    ll t=1;
    if(j*t<x){
        ans="b"+ans;
        t++;
    }
    cout<<ans;
}

int main(){
    // freopen("input.txt","r",stdin);
    // freopen("output.txt","w",stdout);
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t=1;
    cin>>t;
    while(t--){
        solve();
        cout<<'\n';
    }
    return 0;
}

ull binpow(ll a,ll b,ll m){
    if(b==0)return ll(1);
    ull a1=binpow(a,b/2,m);
    if(b%2) {
        a1 = (a1*a1)%m;
        return (a*a1)%m;
    }
    else return (a1*a1)%m;
}

void primesfill(int k){
    primes = vector<int>(k+1,1);
    primes[1]=0;
    for(int i=2;i<=k;i++){
        if(!primes[i])continue;
        for(int j=i*2;j<=k;j+=i){
            primes[j]=0;
        }
    }
}