#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    unsigned long long n, k, x;
    cin >> n >> k >> x;
    string s;
    cin >> s;
    vector<unsigned long long> vec;
    vec.reserve(n);
    unsigned long long cnt = 0;
    for (int i = n - 1; i >= 0; i--) {
      if (s[i] == 'a') {
        if (i < n - 1 && s[i] != s[i + 1]) {
          vec.push_back(cnt * k);
          cnt = 0;
        }
      } else
        cnt++;
    }
    if (cnt > 0) vec.push_back(cnt * k);
    vector<unsigned long long> vec2(vec.size(), 0);
    if (vec.size() > 0) {
      vec2[0] = 1;
      unsigned long long sum = vec2[0] * vec[0];
      for (int i = 1; i < vec.size(); i++) {
        if (sum > 1e18) {
          vec2[i] = 1e19;
        } else {
          vec2[i] = sum + 1;
          sum += vec2[i] * vec[i];
        }
      }
    }
    x--;
    vector<unsigned long long> vec3(vec2.size(), 0);
    for (int i = vec2.size() - 1; i >= 0; i--) {
      if (x >= vec2[i]) {
        vec3[i] = min(vec[i], x / vec2[i]);
        x -= vec3[i] * vec2[i];
      }
    }
    int p = vec3.size() - 1;
    string ans = "";
    for (int i = 0; i < n; i++) {
      if (s[i] == 'a')
        ans += 'a';
      else if (s[i] == '*') {
        if (i == 0 || (i > 0 && s[i - 1] == 'a')) {
          for (int j = 0; j < vec3[p]; j++) ans += 'b';
          p--;
        }
      }
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    unsigned long long n, k, x;
    cin >> n >> k >> x;
    string s;
    cin >> s;
    vector<unsigned long long> vec;
    vec.reserve(n);
    unsigned long long cnt = 0;
    for (int i = n - 1; i >= 0; i--) {
      if (s[i] == 'a') {
        if (i < n - 1 && s[i] != s[i + 1]) {
          vec.push_back(cnt * k);
          cnt = 0;
        }
      } else
        cnt++;
    }
    if (cnt > 0) vec.push_back(cnt * k);
    vector<unsigned long long> vec2(vec.size(), 0);
    if (vec.size() > 0) {
      vec2[0] = 1;
      unsigned long long sum = vec2[0] * vec[0];
      for (int i = 1; i < vec.size(); i++) {
        if (sum > 1e18) {
          vec2[i] = 1e19;
        } else {
          vec2[i] = sum + 1;
          sum += vec2[i] * vec[i];
        }
      }
    }
    x--;
    vector<unsigned long long> vec3(vec2.size(), 0);
    for (int i = vec2.size() - 1; i >= 0; i--) {
      if (x >= vec2[i]) {
        vec3[i] = min(vec[i], x / vec2[i]);
        x -= vec3[i] * vec2[i];
      }
    }
    int p = vec3.size() - 1;
    string ans = "";
    for (int i = 0; i < n; i++) {
      if (s[i] == 'a')
        ans += 'a';
      else if (s[i] == '*') {
        if (i == 0 || (i > 0 && s[i - 1] == 'a')) {
          for (int j = 0; j < vec3[p]; j++) ans += 'b';
          p--;
        }
      }
    }
    cout << ans << endl;
  }
  return 0;
}

```