#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  long long t, n, k, x;
  cin >> t;
  string st;
  while (t--) {
    cin >> n >> k >> x;
    x--;
    cin >> st;
    string s = "";
    vector<long long> v;
    for (long long i = 0; i < st.length();) {
      if (st[i] == 'a') {
        v.push_back(-1);
        i++;
      } else {
        long long c = 0;
        while (i < n && st[i] == '*') i++, c += k;
        v.push_back(c);
      }
    }
    vector<long long> v1(v.size());
    for (long long i = v1.size() - 1; i >= 0; i--) {
      if (v[i] < 0)
        v1[i] = -1;
      else {
        v1[i] = x % (v[i] + 1);
        x = x / (v[i] + 1);
      }
    }
    for (long long i = 0; i < v1.size(); i++) {
      if (v1[i] == 0) continue;
      if (v1[i] == -1)
        s += "a";
      else {
        for (long long j = 0; j < v1[i]; j++) s += "b";
      }
    }
    cout << s << endl;
  }
  return 0;
}

### Prompt

Please formulate 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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  long long t, n, k, x;
  cin >> t;
  string st;
  while (t--) {
    cin >> n >> k >> x;
    x--;
    cin >> st;
    string s = "";
    vector<long long> v;
    for (long long i = 0; i < st.length();) {
      if (st[i] == 'a') {
        v.push_back(-1);
        i++;
      } else {
        long long c = 0;
        while (i < n && st[i] == '*') i++, c += k;
        v.push_back(c);
      }
    }
    vector<long long> v1(v.size());
    for (long long i = v1.size() - 1; i >= 0; i--) {
      if (v[i] < 0)
        v1[i] = -1;
      else {
        v1[i] = x % (v[i] + 1);
        x = x / (v[i] + 1);
      }
    }
    for (long long i = 0; i < v1.size(); i++) {
      if (v1[i] == 0) continue;
      if (v1[i] == -1)
        s += "a";
      else {
        for (long long j = 0; j < v1[i]; j++) s += "b";
      }
    }
    cout << s << endl;
  }
  return 0;
}

```