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

//木の葉舞う所に，火は燃ゆる。
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cstdlib>
#include <iomanip>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <string>
#include <cstring>
#include <vector>
#include <cmath>
#include <bitset>
#include <ext/rope>
#include <unordered_set>
#include <unordered_map>
#define int long long
#define inf 0x3f3f3f3f
#define debug printf("Hello World\n")
#define ios ios::sync_with_stdio(false)
using namespace __gnu_cxx;
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
const int mod1 = 1e9 + 7, mod2 = 998244353;
const double PI = 3.1415926535898, Ee = 2.718281828459;
const int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};
inline int read()
{
    int x = 0, f = 1;char ch = getchar();
    while (!isdigit(ch)) {if (ch == '-') f = -1; ch = getchar();}
    while (isdigit(ch)) {x = x * 10 + ch - 48; ch = getchar();}
    return x * f;
}
const int N = 2000 + 10, M = 1e6 + 10;

int vis[N];

void solve()
{
    int n, k, x; cin>>n>>k>>x;
    for(int i = 0;i <= n;i ++) vis[i] = 0;
    string s, t; cin>>s;

    for(int i = 0;i < s.size();i ++)
    {
        t += s[i];
        if(s[i] == '*')
        {
            int j = i;
            while(j < s.size() && s[j] == '*') j ++;
            vis[(int)t.size() - 1] = (j - i) * k;
            i = j - 1;
        }
    }

    string ans;
    int sum = 1;
    for(int i = (int)t.size() - 1;i >= 0;i --)
    {
        if(t[i] == 'a') continue;

        if(vis[i] + 1 >= (x + sum - 1) / sum)
        {
            for(int j = 0;j < i;j ++) if(t[j] == 'a') ans += t[j];
            for(int j = i;j < t.size();j ++)
            {
                if(t[j] == 'a') ans += t[j];
                else
                {
                    if(j != i) sum /= vis[j] + 1;
                    int tot = (x + sum - 1) / sum - 1;
                    x -= tot * sum;
                    for(int tt = 1;tt <= tot;tt ++) ans += 'b';
                }
            }
            break;
        }
        else sum *= vis[i] + 1;
    }
    cout<<ans<<endl;
}

signed main()
{
    ios;
    int t = 1;  cin>>t;
    while(t --) solve();
    return 0;
}