#include <bits/stdc++.h>
using namespace std;
long long p[4000005];
int sz[4000005];
char s[4000005], t[4000005];
int main() {
  long long x;
  int i, k, tt, n, m, cnt;
  scanf("%d", &tt);
  while (tt--) {
    cnt = 0;
    scanf("%d %d %lld %s", &n, &m, &x, s + 1);
    for (i = 1; i <= n; ++i)
      if (s[i] == 'a')
        t[++cnt] = 'a';
      else if (t[cnt] != '*') {
        t[++cnt] = '*';
        sz[cnt] = m;
      } else
        sz[cnt] += m;
    p[cnt + 1] = 1;
    for (i = n = cnt, cnt = 0; i; --i)
      if (t[i] == 'a')
        p[i] = p[i + 1];
      else {
        if (p[i + 1] == -1 || x / p[i + 1] < (sz[i] + 1))
          p[i] = -1;
        else
          p[i] = p[i + 1] * (sz[i] + 1);
      }
    for (i = 1; i <= n; ++i) {
      if (t[i] == 'a')
        s[++cnt] = 'a';
      else {
        if (p[i + 1] == -1)
          continue;
        else
          for (k = 1; k <= sz[i]; ++k)
            if (x > p[i + 1]) {
              s[++cnt] = 'b';
              x -= p[i + 1];
            }
      }
    }
    s[cnt + 1] = 0;
    puts(s + 1);
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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;
long long p[4000005];
int sz[4000005];
char s[4000005], t[4000005];
int main() {
  long long x;
  int i, k, tt, n, m, cnt;
  scanf("%d", &tt);
  while (tt--) {
    cnt = 0;
    scanf("%d %d %lld %s", &n, &m, &x, s + 1);
    for (i = 1; i <= n; ++i)
      if (s[i] == 'a')
        t[++cnt] = 'a';
      else if (t[cnt] != '*') {
        t[++cnt] = '*';
        sz[cnt] = m;
      } else
        sz[cnt] += m;
    p[cnt + 1] = 1;
    for (i = n = cnt, cnt = 0; i; --i)
      if (t[i] == 'a')
        p[i] = p[i + 1];
      else {
        if (p[i + 1] == -1 || x / p[i + 1] < (sz[i] + 1))
          p[i] = -1;
        else
          p[i] = p[i + 1] * (sz[i] + 1);
      }
    for (i = 1; i <= n; ++i) {
      if (t[i] == 'a')
        s[++cnt] = 'a';
      else {
        if (p[i + 1] == -1)
          continue;
        else
          for (k = 1; k <= sz[i]; ++k)
            if (x > p[i + 1]) {
              s[++cnt] = 'b';
              x -= p[i + 1];
            }
      }
    }
    s[cnt + 1] = 0;
    puts(s + 1);
  }
  return 0;
}

```