#include <bits/stdc++.h>
using namespace std;
string red(string s) {
  string ss = "";
  for (long long i = 0; i < s.length(); i++)
    if (s[i] != '*') ss += s[i];
  return ss;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long tt;
  cin >> tt;
  while (tt--) {
    long long n, k, x;
    cin >> n >> k >> x;
    string s;
    cin >> s;
    if (x == 1) {
      cout << red(s) << "\n";
      continue;
    }
    vector<vector<long long> > v;
    for (long long i = n - 1; i >= 0; i--) {
      if (s[i] == 'a') {
        continue;
      }
      long long j = i;
      while (j >= 0 && s[i] == s[j]) j--;
      v.push_back({i, (i - j) * k + 1});
      i = j + 1;
    }
    long long prd = 1;
    long long i;
    reverse(v.begin(), v.end());
    for (i = (long long)v.size() - 1; i >= 0; i--) {
      if (v[i][1] >= ceil((double)x / prd)) break;
      prd = prd * v[i][1];
    }
    i = max(i, (long long)0);
    string ans = s.substr(0, v[i][0]);
    for (long long j = i; j < v.size(); j++) {
      if (j != i) prd /= v[j][1];
      long long cnt = ceil((double)x / prd);
      for (long long k = 1; k < cnt; k++) ans += 'b';
      x -= (cnt - 1) * (prd);
      if (j == (long long)v.size() - 1) {
        for (long long k = v[j][0] + 1; k < n; k++) ans += s[k];
      } else {
        for (long long k = v[j][0] + 1; k < v[j + 1][0]; k++) ans += s[k];
      }
    }
    ans = red(ans);
    cout << ans << "\n";
  }
}

### 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;
string red(string s) {
  string ss = "";
  for (long long i = 0; i < s.length(); i++)
    if (s[i] != '*') ss += s[i];
  return ss;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long tt;
  cin >> tt;
  while (tt--) {
    long long n, k, x;
    cin >> n >> k >> x;
    string s;
    cin >> s;
    if (x == 1) {
      cout << red(s) << "\n";
      continue;
    }
    vector<vector<long long> > v;
    for (long long i = n - 1; i >= 0; i--) {
      if (s[i] == 'a') {
        continue;
      }
      long long j = i;
      while (j >= 0 && s[i] == s[j]) j--;
      v.push_back({i, (i - j) * k + 1});
      i = j + 1;
    }
    long long prd = 1;
    long long i;
    reverse(v.begin(), v.end());
    for (i = (long long)v.size() - 1; i >= 0; i--) {
      if (v[i][1] >= ceil((double)x / prd)) break;
      prd = prd * v[i][1];
    }
    i = max(i, (long long)0);
    string ans = s.substr(0, v[i][0]);
    for (long long j = i; j < v.size(); j++) {
      if (j != i) prd /= v[j][1];
      long long cnt = ceil((double)x / prd);
      for (long long k = 1; k < cnt; k++) ans += 'b';
      x -= (cnt - 1) * (prd);
      if (j == (long long)v.size() - 1) {
        for (long long k = v[j][0] + 1; k < n; k++) ans += s[k];
      } else {
        for (long long k = v[j][0] + 1; k < v[j + 1][0]; k++) ans += s[k];
      }
    }
    ans = red(ans);
    cout << ans << "\n";
  }
}

```