#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 1e5;
const long long MOD = 1000000007;
long long t, n, m, k, x, y, a, b;
void solve() {
  cin >> n >> k >> x;
  x--;
  string s;
  cin >> s;
  char c = 'a';
  vector<long long> v;
  long long cnt = -1;
  for (int i = 0; i < n; i++) {
    if (s[i] == '*') {
      if (cnt == -1) {
        cnt++;
      }
      cnt += k;
    } else {
      if (cnt != -1) {
        cnt++;
        v.push_back(cnt);
      }
      cnt = -1;
    }
    c = s[i];
  }
  if (cnt != -1) {
    cnt++;
    v.push_back(cnt);
  }
  reverse(v.begin(), v.end());
  for (int i = 0; i < v.size(); i++) {
    long long y = v[i];
    v[i] = x % y;
    x /= y;
  }
  reverse(v.begin(), v.end());
  long long pos = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] == 'a') {
      cout << 'a';
    } else {
      for (int j = 0; j < v[pos]; j++) {
        cout << 'b';
      }
      pos++;
      if (pos >= v.size()) {
        pos--;
      }
      if (i != n - 1) {
        while (i < n - 1 && s[i + 1] == '*') {
          i++;
          if (i == n - 1) {
            break;
          }
        }
      }
    }
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
  cout << setprecision(10);
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Generate 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;
const long long MAXN = 1e5;
const long long MOD = 1000000007;
long long t, n, m, k, x, y, a, b;
void solve() {
  cin >> n >> k >> x;
  x--;
  string s;
  cin >> s;
  char c = 'a';
  vector<long long> v;
  long long cnt = -1;
  for (int i = 0; i < n; i++) {
    if (s[i] == '*') {
      if (cnt == -1) {
        cnt++;
      }
      cnt += k;
    } else {
      if (cnt != -1) {
        cnt++;
        v.push_back(cnt);
      }
      cnt = -1;
    }
    c = s[i];
  }
  if (cnt != -1) {
    cnt++;
    v.push_back(cnt);
  }
  reverse(v.begin(), v.end());
  for (int i = 0; i < v.size(); i++) {
    long long y = v[i];
    v[i] = x % y;
    x /= y;
  }
  reverse(v.begin(), v.end());
  long long pos = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] == 'a') {
      cout << 'a';
    } else {
      for (int j = 0; j < v[pos]; j++) {
        cout << 'b';
      }
      pos++;
      if (pos >= v.size()) {
        pos--;
      }
      if (i != n - 1) {
        while (i < n - 1 && s[i + 1] == '*') {
          i++;
          if (i == n - 1) {
            break;
          }
        }
      }
    }
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
  cout << setprecision(10);
  cin >> t;
  while (t--) {
    solve();
  }
}

```