#include <bits/stdc++.h>
#pragma GCC optimize("O1")
using namespace std;
void solve() {
  long long int n, k, x;
  cin >> n >> k >> x;
  x--;
  string s;
  cin >> s;
  long long int c = 0;
  long long int p = 1;
  vector<pair<char, long long int>> v;
  for (long long int i = 0; i < n; i++) {
    if (s[i] == '*') {
      c += k;
    } else {
      c++;
      if (c > 1) v.push_back({'b', c});
      v.push_back({'a', 1});
      c = 0;
    }
  }
  c++;
  if (c > 1) v.push_back({'b', c});
  reverse(v.begin(), v.end());
  string ans = "";
  long long int l = v.size();
  for (long long int i = 0; i < l; i++) {
    if (v[i].first == 'b') {
      long long int p = x % v[i].second;
      x /= v[i].second;
      for (long long int j = 0; j < p; j++) ans.push_back('b');
    } else
      ans.push_back('a');
  }
  reverse(ans.begin(), ans.end());
  cout << ans;
  cout << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int tt;
  cin >> tt;
  while (tt--) {
    solve();
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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>
#pragma GCC optimize("O1")
using namespace std;
void solve() {
  long long int n, k, x;
  cin >> n >> k >> x;
  x--;
  string s;
  cin >> s;
  long long int c = 0;
  long long int p = 1;
  vector<pair<char, long long int>> v;
  for (long long int i = 0; i < n; i++) {
    if (s[i] == '*') {
      c += k;
    } else {
      c++;
      if (c > 1) v.push_back({'b', c});
      v.push_back({'a', 1});
      c = 0;
    }
  }
  c++;
  if (c > 1) v.push_back({'b', c});
  reverse(v.begin(), v.end());
  string ans = "";
  long long int l = v.size();
  for (long long int i = 0; i < l; i++) {
    if (v[i].first == 'b') {
      long long int p = x % v[i].second;
      x /= v[i].second;
      for (long long int j = 0; j < p; j++) ans.push_back('b');
    } else
      ans.push_back('a');
  }
  reverse(ans.begin(), ans.end());
  cout << ans;
  cout << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int tt;
  cin >> tt;
  while (tt--) {
    solve();
  }
  return 0;
}

```