#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  ll T;
  cin >> T;
  while (T--) {
    ll x;
    ll n, k;
    cin >> n >> k >> x;
    x--;
    string S;
    cin >> S;
    vector<ll> star_pos;
    vector<ll> vals;
    for (ll i = 0; i < n; i++) {
      if (S[i] == '*') {
        ll now = i;
        star_pos.push_back(i);
        while (S[i] == '*') {
          i++;
        }
        vals.push_back((i - now) * k + 1);
        i--;
      }
    }
    vector<ll> dig;
    dig.resize(vals.size());
    ll ed = vals.size() - 1;
    while (x) {
      dig[ed] = (x % vals[ed]);
      x /= vals[ed];
      ed -= 1;
      if (ed < 0) break;
    }
    ll ptrs = 0;
    for (ll i = 0; i < n; i++) {
      if (S[i] == '*') {
        for (ll j = 0; j < dig[ptrs]; j++) {
          cout << 'b';
        }
        ptrs += 1;
        while (S[i] == '*') {
          i++;
        }
        i--;
      } else {
        cout << S[i];
      }
    }
    cout << '\n';
  }
}

### Prompt

Create a solution in Cpp for the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  ll T;
  cin >> T;
  while (T--) {
    ll x;
    ll n, k;
    cin >> n >> k >> x;
    x--;
    string S;
    cin >> S;
    vector<ll> star_pos;
    vector<ll> vals;
    for (ll i = 0; i < n; i++) {
      if (S[i] == '*') {
        ll now = i;
        star_pos.push_back(i);
        while (S[i] == '*') {
          i++;
        }
        vals.push_back((i - now) * k + 1);
        i--;
      }
    }
    vector<ll> dig;
    dig.resize(vals.size());
    ll ed = vals.size() - 1;
    while (x) {
      dig[ed] = (x % vals[ed]);
      x /= vals[ed];
      ed -= 1;
      if (ed < 0) break;
    }
    ll ptrs = 0;
    for (ll i = 0; i < n; i++) {
      if (S[i] == '*') {
        for (ll j = 0; j < dig[ptrs]; j++) {
          cout << 'b';
        }
        ptrs += 1;
        while (S[i] == '*') {
          i++;
        }
        i--;
      } else {
        cout << S[i];
      }
    }
    cout << '\n';
  }
}

```