#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
const int mod = 7 + 1e9;
inline int add(int a, int b) { return a + b >= mod ? a + b - mod : a + b; }
inline int mul(int a, int b) { return a * 1ll * b % mod; }
void wabalabadubdub() {
  long long n, k, x;
  string s, res;
  cin >> n >> k >> x >> s;
  --x;
  for (int i = s.size() - 1; i >= 0; --i) {
    if (s[i] == 'a') {
      res.push_back('a');
    } else {
      int j = i + 1;
      while (j > 0 && s[j - 1] == s[i]) --j;
      int b = x % ((i - j + 1) * k + 1);
      x /= ((i - j + 1) * k + 1);
      while (b--) res.push_back('b');
      i = j;
    }
  }
  reverse(res.begin(), res.end());
  cout << res << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) wabalabadubdub();
}

### Prompt

Generate a Cpp solution to 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;
const int N = 1e6 + 10;
const int mod = 7 + 1e9;
inline int add(int a, int b) { return a + b >= mod ? a + b - mod : a + b; }
inline int mul(int a, int b) { return a * 1ll * b % mod; }
void wabalabadubdub() {
  long long n, k, x;
  string s, res;
  cin >> n >> k >> x >> s;
  --x;
  for (int i = s.size() - 1; i >= 0; --i) {
    if (s[i] == 'a') {
      res.push_back('a');
    } else {
      int j = i + 1;
      while (j > 0 && s[j - 1] == s[i]) --j;
      int b = x % ((i - j + 1) * k + 1);
      x /= ((i - j + 1) * k + 1);
      while (b--) res.push_back('b');
      i = j;
    }
  }
  reverse(res.begin(), res.end());
  cout << res << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) wabalabadubdub();
}

```