#include <bits/stdc++.h>
using namespace std;
string s;
int n, k, t, cnt[2001];
unsigned long long x;
int main() {
  cin >> t;
  while (t--) {
    cin >> n >> k >> x;
    cin >> s;
    string news = "";
    int len = -1;
    for (int i = 0; i < n; i++) {
      if (i == 0) {
        news += s[i];
        if (s[i] == '*') {
          len++;
          cnt[len] = k;
        } else {
          len++;
          cnt[len] = 0;
        }
      } else {
        if (s[i] == s[i - 1] && s[i] == '*') {
          cnt[len] += k;
        } else {
          news += s[i];
          if (s[i] == '*') {
            len++;
            cnt[len] = k;
          } else {
            len++;
            cnt[len] = 0;
          }
        }
      }
    }
    len = news.size();
    unsigned long long tt = 1, vt = 0;
    vector<unsigned long long> thutu;
    thutu.push_back(1);
    for (int i = len - 1; i >= 0; i--) {
      unsigned long long prett = tt;
      if (cnt[i] > 0) {
        tt = tt * (cnt[i] + 1);
        thutu.push_back(tt);
        if (x / prett <= cnt[i]) {
          vt = i;
          break;
        }
      }
    }
    for (int i = 0; i < vt; i++) {
      if (news[i] == 'a') {
        cout << news[i];
      }
    }
    if (thutu.size()) {
      thutu.pop_back();
    }
    for (int i = vt; i < len; i++) {
      if (news[i] == 'a') {
        cout << news[i];
      } else {
        if (thutu.size()) {
          unsigned long long giasu = thutu.back();
          thutu.pop_back();
          unsigned long long o = x / giasu;
          if (x % giasu == 0) {
            o--;
          }
          x -= o * giasu;
          for (int j = 1; j <= o; j++) {
            cout << 'b';
          }
        }
      }
    }
    cout << "\n";
  }
}

### Prompt

Please formulate a Cpp solution 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;
string s;
int n, k, t, cnt[2001];
unsigned long long x;
int main() {
  cin >> t;
  while (t--) {
    cin >> n >> k >> x;
    cin >> s;
    string news = "";
    int len = -1;
    for (int i = 0; i < n; i++) {
      if (i == 0) {
        news += s[i];
        if (s[i] == '*') {
          len++;
          cnt[len] = k;
        } else {
          len++;
          cnt[len] = 0;
        }
      } else {
        if (s[i] == s[i - 1] && s[i] == '*') {
          cnt[len] += k;
        } else {
          news += s[i];
          if (s[i] == '*') {
            len++;
            cnt[len] = k;
          } else {
            len++;
            cnt[len] = 0;
          }
        }
      }
    }
    len = news.size();
    unsigned long long tt = 1, vt = 0;
    vector<unsigned long long> thutu;
    thutu.push_back(1);
    for (int i = len - 1; i >= 0; i--) {
      unsigned long long prett = tt;
      if (cnt[i] > 0) {
        tt = tt * (cnt[i] + 1);
        thutu.push_back(tt);
        if (x / prett <= cnt[i]) {
          vt = i;
          break;
        }
      }
    }
    for (int i = 0; i < vt; i++) {
      if (news[i] == 'a') {
        cout << news[i];
      }
    }
    if (thutu.size()) {
      thutu.pop_back();
    }
    for (int i = vt; i < len; i++) {
      if (news[i] == 'a') {
        cout << news[i];
      } else {
        if (thutu.size()) {
          unsigned long long giasu = thutu.back();
          thutu.pop_back();
          unsigned long long o = x / giasu;
          if (x % giasu == 0) {
            o--;
          }
          x -= o * giasu;
          for (int j = 1; j <= o; j++) {
            cout << 'b';
          }
        }
      }
    }
    cout << "\n";
  }
}

```