#include <bits/stdc++.h>
using namespace std;
const long long N = 2010;
long long T;
long long a[N];
long long n, k, x;
char s[N];
signed main() {
  scanf("%lld", &T);
  while (T--) {
    scanf("%lld%lld%lld%s", &n, &k, &x, s + 1);
    long long cnt = 0, sum = 0;
    for (long long i = 1; i <= n; i++) {
      if (s[i] == '*')
        sum++;
      else if (sum)
        a[++cnt] = sum * k + 1, sum = 0;
    }
    if (sum) a[++cnt] = sum * k + 1, sum = 0;
    long double P = 1.0;
    long long L = 0;
    for (long long i = cnt; i >= 1; i--) {
      P *= a[i];
      if (P >= x) {
        L = i;
        break;
      }
    }
    cnt = sum = 0;
    for (long long i = 1; i <= n; i++) {
      if (s[i] == '*')
        sum++;
      else {
        if (sum && ++cnt >= L) {
          assert(a[cnt]);
          P /= a[cnt];
          for (long long k = 1; k < (x - 1) / ((long long)P) + 1; k++)
            putchar('b');
          x -= (x - 1) / ((long long)P) * ((long long)P);
        }
        putchar('a'), sum = 0;
      }
    }
    if (sum && ++cnt >= L) {
      P /= a[cnt];
      for (long long k = 1; k < (x - 1) / ((long long)P) + 1; k++) putchar('b');
      x -= (x - 1) / ((long long)P) * ((long long)P);
    }
    puts("");
  }
  return 0;
}

### 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;
const long long N = 2010;
long long T;
long long a[N];
long long n, k, x;
char s[N];
signed main() {
  scanf("%lld", &T);
  while (T--) {
    scanf("%lld%lld%lld%s", &n, &k, &x, s + 1);
    long long cnt = 0, sum = 0;
    for (long long i = 1; i <= n; i++) {
      if (s[i] == '*')
        sum++;
      else if (sum)
        a[++cnt] = sum * k + 1, sum = 0;
    }
    if (sum) a[++cnt] = sum * k + 1, sum = 0;
    long double P = 1.0;
    long long L = 0;
    for (long long i = cnt; i >= 1; i--) {
      P *= a[i];
      if (P >= x) {
        L = i;
        break;
      }
    }
    cnt = sum = 0;
    for (long long i = 1; i <= n; i++) {
      if (s[i] == '*')
        sum++;
      else {
        if (sum && ++cnt >= L) {
          assert(a[cnt]);
          P /= a[cnt];
          for (long long k = 1; k < (x - 1) / ((long long)P) + 1; k++)
            putchar('b');
          x -= (x - 1) / ((long long)P) * ((long long)P);
        }
        putchar('a'), sum = 0;
      }
    }
    if (sum && ++cnt >= L) {
      P /= a[cnt];
      for (long long k = 1; k < (x - 1) / ((long long)P) + 1; k++) putchar('b');
      x -= (x - 1) / ((long long)P) * ((long long)P);
    }
    puts("");
  }
  return 0;
}

```