#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  srand(time(NULL));
  ;
  int T = 1;
  cin >> T;
  while (T--) {
    int n, k;
    long long int x;
    cin >> n >> k >> x;
    string s;
    cin >> s;
    vector<long long int> v;
    for (int i = 0; i < int(n); i++) {
      if (s[i] == 'a')
        v.push_back(-1);
      else {
        int j = i + 1;
        while (j < n && s[j] != 'a') j++;
        v.push_back(1ll * k * (j - i));
        i = j - 1;
      }
    }
    n = int((v).size());
    int j = n;
    long long int m = 1, xx = x - 1;
    for (int i = n - 1; i >= 0; i--)
      if (v[i] > 0) {
        if (xx < v[i] + 1) {
          j = i;
          break;
        } else
          xx /= (v[i] + 1), m *= v[i] + 1;
      }
    for (int i = 0; i < int(j); i++)
      if (v[i] > 0) v[i] = 0;
    if (j < n) v[j] = (x - 1) / m;
    long long int y = (x - 1) % m + 1;
    for (int i = j + 1; i <= n - 1; i++)
      if (v[i] > 0) {
        m /= v[i] + 1;
        v[i] = (y - 1) / m;
        y = (y - 1) % m + 1;
      }
    for (int i = 0; i < int(n); i++) {
      if (v[i] < 0)
        cout << "a";
      else
        for (long long int z = 0; z < v[i]; z++) cout << "b";
    }
    cout << endl;
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  srand(time(NULL));
  ;
  int T = 1;
  cin >> T;
  while (T--) {
    int n, k;
    long long int x;
    cin >> n >> k >> x;
    string s;
    cin >> s;
    vector<long long int> v;
    for (int i = 0; i < int(n); i++) {
      if (s[i] == 'a')
        v.push_back(-1);
      else {
        int j = i + 1;
        while (j < n && s[j] != 'a') j++;
        v.push_back(1ll * k * (j - i));
        i = j - 1;
      }
    }
    n = int((v).size());
    int j = n;
    long long int m = 1, xx = x - 1;
    for (int i = n - 1; i >= 0; i--)
      if (v[i] > 0) {
        if (xx < v[i] + 1) {
          j = i;
          break;
        } else
          xx /= (v[i] + 1), m *= v[i] + 1;
      }
    for (int i = 0; i < int(j); i++)
      if (v[i] > 0) v[i] = 0;
    if (j < n) v[j] = (x - 1) / m;
    long long int y = (x - 1) % m + 1;
    for (int i = j + 1; i <= n - 1; i++)
      if (v[i] > 0) {
        m /= v[i] + 1;
        v[i] = (y - 1) / m;
        y = (y - 1) % m + 1;
      }
    for (int i = 0; i < int(n); i++) {
      if (v[i] < 0)
        cout << "a";
      else
        for (long long int z = 0; z < v[i]; z++) cout << "b";
    }
    cout << endl;
  }
  return 0;
}

```