#include <bits/stdc++.h>
using namespace std;
long long int mul(long long int a, long long int b) {
  long long int ans = 0;
  while (b) {
    if (b & 1) {
      ans += a;
    }
    b = b >> 1;
    a += a;
    if (a > 1e18) {
      a = 1e18;
    }
    if (ans > 1e18) {
      ans = 1e18;
    }
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long int i, j, k, l, m, n, a, b, c, d, u, v, w, t = 1, x;
  cin >> t;
  while (t--) {
    long long int n, k, x;
    cin >> n >> k >> x;
    x--;
    string s;
    cin >> s;
    vector<long long int> cnt;
    long long int c = 0;
    for (long long int i = 0; i < n; ++i) {
      if (s[i] == 'a') {
        cnt.push_back(c);
        c = 0;
      } else {
        c++;
      }
    }
    cnt.push_back(c);
    vector<long long int> m(cnt.size());
    m[cnt.size() - 1] = 1;
    for (long long int i = cnt.size() - 2; i >= 0; --i) {
      m[i] = mul(m[i + 1], (k * cnt[i + 1] + 1));
    }
    vector<long long int> ans(cnt.size());
    for (long long int i = 0; i < cnt.size(); ++i) {
      ans[i] = x / m[i];
      x %= m[i];
    }
    for (long long int i = 0; i < cnt.size(); ++i) {
      for (long long int j = 0; j < ans[i]; ++j) {
        cout << 'b';
      }
      if (i + 1 != cnt.size()) {
        cout << 'a';
      }
    }
    cout << endl;
  }
}

### Prompt

Your task is to create 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;
long long int mul(long long int a, long long int b) {
  long long int ans = 0;
  while (b) {
    if (b & 1) {
      ans += a;
    }
    b = b >> 1;
    a += a;
    if (a > 1e18) {
      a = 1e18;
    }
    if (ans > 1e18) {
      ans = 1e18;
    }
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long int i, j, k, l, m, n, a, b, c, d, u, v, w, t = 1, x;
  cin >> t;
  while (t--) {
    long long int n, k, x;
    cin >> n >> k >> x;
    x--;
    string s;
    cin >> s;
    vector<long long int> cnt;
    long long int c = 0;
    for (long long int i = 0; i < n; ++i) {
      if (s[i] == 'a') {
        cnt.push_back(c);
        c = 0;
      } else {
        c++;
      }
    }
    cnt.push_back(c);
    vector<long long int> m(cnt.size());
    m[cnt.size() - 1] = 1;
    for (long long int i = cnt.size() - 2; i >= 0; --i) {
      m[i] = mul(m[i + 1], (k * cnt[i + 1] + 1));
    }
    vector<long long int> ans(cnt.size());
    for (long long int i = 0; i < cnt.size(); ++i) {
      ans[i] = x / m[i];
      x %= m[i];
    }
    for (long long int i = 0; i < cnt.size(); ++i) {
      for (long long int j = 0; j < ans[i]; ++j) {
        cout << 'b';
      }
      if (i + 1 != cnt.size()) {
        cout << 'a';
      }
    }
    cout << endl;
  }
}

```