#include <bits/stdc++.h>
using namespace std;
const int N = 3500;
char s[N];
unsigned long long d[N];
unsigned long long id[N];
unsigned long long tx[N];
int main() {
  int T;
  cin >> T;
  while (T--) {
    memset(tx, 0, sizeof tx);
    memset(d, 0, sizeof d);
    memset(id, 0, sizeof id);
    int n, k;
    cin >> n >> k;
    unsigned long long x;
    cin >> x;
    cin >> s;
    x--;
    bool falg = false;
    if (k == 0) {
      for (int i = 0; i < n; i++) {
        if (s[i] == 'a') cout << s[i];
      }
      cout << endl;
      continue;
    }
    unsigned long long cnt = 0;
    unsigned long long j = 0;
    for (int i = 0; i <= n - 1; i++) {
      if (s[i] == 'a' && cnt != 0) {
        id[j] = i;
        d[j++] = cnt;
        cnt = 0;
      } else if (s[i] == '*') {
        cnt += k;
      }
    }
    if (cnt) id[j] = n, d[j++] = cnt;
    unsigned long long last = 1;
    int m = -1;
    for (int i = j - 1; i >= 0; i--) {
      if (d[i] * last >= x) {
        m = i;
        tx[i] = x / last;
        x -= last * tx[i];
        break;
      } else
        last = (last) * (d[i] + 1);
    }
    for (int i = m + 1; i < j; i++) {
      if (!x) break;
      last /= (d[i] + 1);
      tx[i] = x / last;
      x -= last * tx[i];
    }
    int tt = 0;
    for (int i = 0; i < n; i++) {
      if (s[i] == 'a') {
        cout << "a";
      } else {
        for (int z = 0; z < tx[tt]; z++) cout << "b";
        i = id[tt] - 1;
        tt++;
      }
    }
    cout << endl;
  }
}

### Prompt

Your task is to create 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;
const int N = 3500;
char s[N];
unsigned long long d[N];
unsigned long long id[N];
unsigned long long tx[N];
int main() {
  int T;
  cin >> T;
  while (T--) {
    memset(tx, 0, sizeof tx);
    memset(d, 0, sizeof d);
    memset(id, 0, sizeof id);
    int n, k;
    cin >> n >> k;
    unsigned long long x;
    cin >> x;
    cin >> s;
    x--;
    bool falg = false;
    if (k == 0) {
      for (int i = 0; i < n; i++) {
        if (s[i] == 'a') cout << s[i];
      }
      cout << endl;
      continue;
    }
    unsigned long long cnt = 0;
    unsigned long long j = 0;
    for (int i = 0; i <= n - 1; i++) {
      if (s[i] == 'a' && cnt != 0) {
        id[j] = i;
        d[j++] = cnt;
        cnt = 0;
      } else if (s[i] == '*') {
        cnt += k;
      }
    }
    if (cnt) id[j] = n, d[j++] = cnt;
    unsigned long long last = 1;
    int m = -1;
    for (int i = j - 1; i >= 0; i--) {
      if (d[i] * last >= x) {
        m = i;
        tx[i] = x / last;
        x -= last * tx[i];
        break;
      } else
        last = (last) * (d[i] + 1);
    }
    for (int i = m + 1; i < j; i++) {
      if (!x) break;
      last /= (d[i] + 1);
      tx[i] = x / last;
      x -= last * tx[i];
    }
    int tt = 0;
    for (int i = 0; i < n; i++) {
      if (s[i] == 'a') {
        cout << "a";
      } else {
        for (int z = 0; z < tx[tt]; z++) cout << "b";
        i = id[tt] - 1;
        tt++;
      }
    }
    cout << endl;
  }
}

```