#include <bits/stdc++.h>
using namespace std;
using ull = unsigned long long;
int main() {
  int t;
  cin >> t;
  while (t--) {
    ull n, k, x;
    cin >> n >> k >> x;
    cin.ignore();
    string arr;
    getline(cin, arr);
    vector<pair<ull, ull>> values;
    for (int i = arr.size(); i-- > 0;) {
      if (arr[i] == 'a') {
        if (i == arr.size() - 1 || arr[i + 1] != 'a')
          values.emplace_back(0, 1);
        else
          values.back().second++;
        continue;
      }
      if (i == arr.size() - 1)
        values.emplace_back(1, k);
      else if (arr[i + 1] == '*')
        values.back().second += k;
      else
        values.emplace_back(1, k);
    }
    x--;
    for (int i = values.size(); i-- > 0;) {
      if (values[i].first == 0) {
        for (int j = 0; j < values[i].second; j++) cout << 'a';
        continue;
      }
      ull weight = 1;
      for (int j = 0; j < values.size(); j++) {
        if (j == i) break;
        if (!values[j].first) continue;
        ull temp = weight * (values[j].second + 1);
        if (temp < weight) {
          weight = 0;
          break;
        }
        weight = temp;
      }
      if (!weight) continue;
      int total_b = x / weight;
      for (int j = 0; j < total_b; j++) cout << 'b';
      x -= total_b * weight;
    }
    cout << endl;
  }
}

### 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;
using ull = unsigned long long;
int main() {
  int t;
  cin >> t;
  while (t--) {
    ull n, k, x;
    cin >> n >> k >> x;
    cin.ignore();
    string arr;
    getline(cin, arr);
    vector<pair<ull, ull>> values;
    for (int i = arr.size(); i-- > 0;) {
      if (arr[i] == 'a') {
        if (i == arr.size() - 1 || arr[i + 1] != 'a')
          values.emplace_back(0, 1);
        else
          values.back().second++;
        continue;
      }
      if (i == arr.size() - 1)
        values.emplace_back(1, k);
      else if (arr[i + 1] == '*')
        values.back().second += k;
      else
        values.emplace_back(1, k);
    }
    x--;
    for (int i = values.size(); i-- > 0;) {
      if (values[i].first == 0) {
        for (int j = 0; j < values[i].second; j++) cout << 'a';
        continue;
      }
      ull weight = 1;
      for (int j = 0; j < values.size(); j++) {
        if (j == i) break;
        if (!values[j].first) continue;
        ull temp = weight * (values[j].second + 1);
        if (temp < weight) {
          weight = 0;
          break;
        }
        weight = temp;
      }
      if (!weight) continue;
      int total_b = x / weight;
      for (int j = 0; j < total_b; j++) cout << 'b';
      x -= total_b * weight;
    }
    cout << endl;
  }
}

```