#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n, k, x;
  cin >> n >> k >> x;
  x--;
  string s;
  cin >> s;
  vector<long long> v;
  bool started = 0;
  long long curcnt = 0;
  for (int i = 0; i < n; i++) {
    if (!started && s[i] == '*') {
      started = 1;
    }
    if (s[i] == '*') {
      curcnt++;
    } else {
      if (curcnt > 0) v.push_back(curcnt * k + 1);
      started = 0;
      curcnt = 0;
    }
  }
  if (curcnt > 0) v.push_back(curcnt * k + 1);
  if (v.size() == 0) {
    cout << s << endl;
    return;
  }
  vector<long long> mod;
  long long prod = 1;
  for (int i = int(v.size() - 1); i >= 0; i--) {
    mod.push_back(x % v[i]);
    x /= v[i];
  }
  reverse(mod.begin(), mod.end());
  int pt = 0;
  bool outputting = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] == 'a') {
      outputting = 0;
      cout << 'a';
      continue;
    } else {
      if (!outputting) {
        for (int j = 0; j < mod[pt]; j++) cout << 'b';
        outputting = 1;
        pt++;
      }
    }
  }
  cout << endl;
  return;
}
int main() {
  int t;
  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;
void solve() {
  long long n, k, x;
  cin >> n >> k >> x;
  x--;
  string s;
  cin >> s;
  vector<long long> v;
  bool started = 0;
  long long curcnt = 0;
  for (int i = 0; i < n; i++) {
    if (!started && s[i] == '*') {
      started = 1;
    }
    if (s[i] == '*') {
      curcnt++;
    } else {
      if (curcnt > 0) v.push_back(curcnt * k + 1);
      started = 0;
      curcnt = 0;
    }
  }
  if (curcnt > 0) v.push_back(curcnt * k + 1);
  if (v.size() == 0) {
    cout << s << endl;
    return;
  }
  vector<long long> mod;
  long long prod = 1;
  for (int i = int(v.size() - 1); i >= 0; i--) {
    mod.push_back(x % v[i]);
    x /= v[i];
  }
  reverse(mod.begin(), mod.end());
  int pt = 0;
  bool outputting = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] == 'a') {
      outputting = 0;
      cout << 'a';
      continue;
    } else {
      if (!outputting) {
        for (int j = 0; j < mod[pt]; j++) cout << 'b';
        outputting = 1;
        pt++;
      }
    }
  }
  cout << endl;
  return;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```