#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    long long x;
    string s;
    cin >> n >> k >> x >> s;
    if (x == 1) {
      for (int i = 0; i < n; ++i) {
        if (s[i] == 'a') {
          cout << s[i];
        }
      }
      cout << "\n";
      continue;
    }
    x--;
    vector<int> cp;
    for (int i = 0; i < n; ++i) {
      if (s[i] == 'a') {
        cp.push_back(i);
      }
    }
    vector<long long> ix;
    for (int i = 0; i < cp.size(); ++i) {
      int sta;
      if (i == 0) {
        sta = -1;
      } else {
        sta = cp[i - 1];
      }
      ix.push_back((cp[i] - sta - 1) * 1LL * k + 1);
    }
    int ed = n - 1 - cp[cp.size() - 1];
    ix.push_back(ed * 1LL * k + 1);
    int op = ix.size();
    vector<long long> ty(op);
    reverse(ix.begin(), ix.end());
    for (int i = 0; i < op; ++i) {
      ty[i] = x % ix[i];
      x /= ix[i];
    }
    string ans;
    for (int i = 0; i + 1 < ty.size(); ++i) {
      for (int j = 0; j < ty[i]; ++j) {
        ans.push_back('b');
      }
      ans.push_back('a');
    }
    for (int i = 0; i < ty[op - 1]; ++i) {
      ans.push_back('b');
    }
    reverse(ans.begin(), ans.end());
    cout << ans << "\n";
  }
  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);
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    long long x;
    string s;
    cin >> n >> k >> x >> s;
    if (x == 1) {
      for (int i = 0; i < n; ++i) {
        if (s[i] == 'a') {
          cout << s[i];
        }
      }
      cout << "\n";
      continue;
    }
    x--;
    vector<int> cp;
    for (int i = 0; i < n; ++i) {
      if (s[i] == 'a') {
        cp.push_back(i);
      }
    }
    vector<long long> ix;
    for (int i = 0; i < cp.size(); ++i) {
      int sta;
      if (i == 0) {
        sta = -1;
      } else {
        sta = cp[i - 1];
      }
      ix.push_back((cp[i] - sta - 1) * 1LL * k + 1);
    }
    int ed = n - 1 - cp[cp.size() - 1];
    ix.push_back(ed * 1LL * k + 1);
    int op = ix.size();
    vector<long long> ty(op);
    reverse(ix.begin(), ix.end());
    for (int i = 0; i < op; ++i) {
      ty[i] = x % ix[i];
      x /= ix[i];
    }
    string ans;
    for (int i = 0; i + 1 < ty.size(); ++i) {
      for (int j = 0; j < ty[i]; ++j) {
        ans.push_back('b');
      }
      ans.push_back('a');
    }
    for (int i = 0; i < ty[op - 1]; ++i) {
      ans.push_back('b');
    }
    reverse(ans.begin(), ans.end());
    cout << ans << "\n";
  }
  return 0;
}

```