#include <bits/stdc++.h>
using namespace std;
const int N = 2e3 + 10;
long long n, k, x;
string str;
long long a[N], b[N];
int idx, idx2;
void calc(long long n, long long k) {
  for (int i = n; i >= 0; i--) {
    b[i] = k % (a[i] + 1);
    k /= (a[i] + 1);
  }
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    cin >> n >> k >> x;
    cin >> str;
    idx = idx2 = 0;
    memset(a, 0, sizeof a);
    memset(b, 0, sizeof b);
    for (int i = 0; i < n; i++) {
      if (str[i] == 'a') {
        if (a[idx]) idx++;
      } else
        a[idx] += k;
    }
    calc(idx, x - 1);
    for (int i = 0; i < n; i++) {
      if (str[i] == 'a') {
        if (i != 0 && str[i - 1] == '*') {
          for (int i = 0; i < b[idx2]; i++) cout << 'b';
          idx2++;
        }
        cout << 'a';
      }
    }
    if (idx2 <= idx)
      while (idx2 <= idx) {
        for (int i = 0; i < b[idx2]; i++) cout << 'b';
        idx2++;
      }
    cout << endl;
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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 = 2e3 + 10;
long long n, k, x;
string str;
long long a[N], b[N];
int idx, idx2;
void calc(long long n, long long k) {
  for (int i = n; i >= 0; i--) {
    b[i] = k % (a[i] + 1);
    k /= (a[i] + 1);
  }
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    cin >> n >> k >> x;
    cin >> str;
    idx = idx2 = 0;
    memset(a, 0, sizeof a);
    memset(b, 0, sizeof b);
    for (int i = 0; i < n; i++) {
      if (str[i] == 'a') {
        if (a[idx]) idx++;
      } else
        a[idx] += k;
    }
    calc(idx, x - 1);
    for (int i = 0; i < n; i++) {
      if (str[i] == 'a') {
        if (i != 0 && str[i - 1] == '*') {
          for (int i = 0; i < b[idx2]; i++) cout << 'b';
          idx2++;
        }
        cout << 'a';
      }
    }
    if (idx2 <= idx)
      while (idx2 <= idx) {
        for (int i = 0; i < b[idx2]; i++) cout << 'b';
        idx2++;
      }
    cout << endl;
  }
  return 0;
}

```