#include <bits/stdc++.h>
using namespace std;
vector<char> vp(4000005);
void ans(vector<long long int> &dp, vector<long long int> &star,
         long long int x, vector<char> &vp, long long int n) {
  if (x <= 1) {
    return;
  }
  long long int starpos = upper_bound((dp).begin(), (dp).end(), x) - dp.begin();
  x -= dp[starpos - 1];
  if (x == 0) {
    for (long long int i = star[starpos - 1]; i < n; i++) {
      if (vp[i] == '*') {
        vp[i] = 'b';
      }
    }
  } else {
    for (long long int i = star[starpos]; i < n; i++) {
      while (i < n && vp[i] == '*') {
        vp[i] = 'b';
        i++;
      }
      break;
    }
    ans(dp, star, x, vp, n);
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int T = 1;
  cin >> T;
  while (T--) {
    long long int n, k, x;
    cin >> n >> k >> x;
    string s;
    cin >> s;
    string b = "";
    for (long long int i = 0; i < k; i++) {
      b += 'b';
    }
    long long int cnt = 0;
    for (long long int i = 0; i < n; i++) {
      if (s[i] == '*') {
        long long int t = cnt;
        for (long long int j = cnt; j < t + k; j++) {
          vp[cnt] = '*';
          cnt++;
        }
      } else {
        vp[cnt] = 'a';
        cnt++;
      }
    }
    s.clear();
    vector<long long int> dp;
    vector<long long int> star;
    star.push_back(-1);
    dp.push_back(1);
    long long int prev = 1;
    long long int curr = 1;
    long long int flag1 = 1;
    for (long long int i = cnt - 1; i >= 0 && flag1; i--) {
      long long int flag = 0;
      while (i >= 0 && vp[i] == '*') {
        if (x < (curr + 1) * 1ll * prev) {
          dp.push_back((curr + 1) * 1ll * prev);
          star.push_back(i);
          flag1 = 0;
          break;
        }
        dp.push_back((curr + 1) * 1ll * prev);
        star.push_back(i);
        i--;
        curr++;
        flag = 1;
      }
      if (flag) {
        prev = curr * prev;
        curr = 1;
      }
    }
    long long int flag = 1;
    ans(dp, star, x, vp, cnt);
    for (long long int i = 0; i < cnt; i++) {
      if (vp[i] == '*') continue;
      cout << vp[i];
    }
    cout << "\n";
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
vector<char> vp(4000005);
void ans(vector<long long int> &dp, vector<long long int> &star,
         long long int x, vector<char> &vp, long long int n) {
  if (x <= 1) {
    return;
  }
  long long int starpos = upper_bound((dp).begin(), (dp).end(), x) - dp.begin();
  x -= dp[starpos - 1];
  if (x == 0) {
    for (long long int i = star[starpos - 1]; i < n; i++) {
      if (vp[i] == '*') {
        vp[i] = 'b';
      }
    }
  } else {
    for (long long int i = star[starpos]; i < n; i++) {
      while (i < n && vp[i] == '*') {
        vp[i] = 'b';
        i++;
      }
      break;
    }
    ans(dp, star, x, vp, n);
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int T = 1;
  cin >> T;
  while (T--) {
    long long int n, k, x;
    cin >> n >> k >> x;
    string s;
    cin >> s;
    string b = "";
    for (long long int i = 0; i < k; i++) {
      b += 'b';
    }
    long long int cnt = 0;
    for (long long int i = 0; i < n; i++) {
      if (s[i] == '*') {
        long long int t = cnt;
        for (long long int j = cnt; j < t + k; j++) {
          vp[cnt] = '*';
          cnt++;
        }
      } else {
        vp[cnt] = 'a';
        cnt++;
      }
    }
    s.clear();
    vector<long long int> dp;
    vector<long long int> star;
    star.push_back(-1);
    dp.push_back(1);
    long long int prev = 1;
    long long int curr = 1;
    long long int flag1 = 1;
    for (long long int i = cnt - 1; i >= 0 && flag1; i--) {
      long long int flag = 0;
      while (i >= 0 && vp[i] == '*') {
        if (x < (curr + 1) * 1ll * prev) {
          dp.push_back((curr + 1) * 1ll * prev);
          star.push_back(i);
          flag1 = 0;
          break;
        }
        dp.push_back((curr + 1) * 1ll * prev);
        star.push_back(i);
        i--;
        curr++;
        flag = 1;
      }
      if (flag) {
        prev = curr * prev;
        curr = 1;
      }
    }
    long long int flag = 1;
    ans(dp, star, x, vp, cnt);
    for (long long int i = 0; i < cnt; i++) {
      if (vp[i] == '*') continue;
      cout << vp[i];
    }
    cout << "\n";
  }
  return 0;
}

```