#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  char ch = getchar(), fl = 1;
  while (ch > '9' || ch < '0') {
    if (ch == '-') fl = -1;
    ch = getchar();
  }
  long long p = 0;
  while (ch >= '0' && ch <= '9') {
    p = p * 10 + ch - '0';
    ch = getchar();
  }
  return p * fl;
}
char ch[2005];
long long len[2005], ans[2005];
int main() {
  int T = read();
  while (T--) {
    long long n = read(), k = read(), x = read() - 1;
    scanf("%s", ch + 1);
    int l = strlen(ch + 1), cnt = 0, num = 0;
    for (int i = 1; i <= l + 1; ++i) {
      if (ch[i] == '*') num++;
      if (ch[i] != '*' && ch[i - 1] == '*') {
        len[++cnt] = k * num + 1;
        num = 0;
      }
    }
    for (int i = cnt; i >= 1; --i) {
      ans[i] = x % len[i];
      x /= len[i];
    }
    num = 1;
    for (int i = 1; i <= n; ++i) {
      if (ch[i] == 'a')
        putchar('a');
      else {
        while (ch[i] == '*') i++;
        --i;
        for (int j = 1; j <= ans[num]; ++j) putchar('b');
        num++;
      }
    }
    putchar('\n');
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
inline long long read() {
  char ch = getchar(), fl = 1;
  while (ch > '9' || ch < '0') {
    if (ch == '-') fl = -1;
    ch = getchar();
  }
  long long p = 0;
  while (ch >= '0' && ch <= '9') {
    p = p * 10 + ch - '0';
    ch = getchar();
  }
  return p * fl;
}
char ch[2005];
long long len[2005], ans[2005];
int main() {
  int T = read();
  while (T--) {
    long long n = read(), k = read(), x = read() - 1;
    scanf("%s", ch + 1);
    int l = strlen(ch + 1), cnt = 0, num = 0;
    for (int i = 1; i <= l + 1; ++i) {
      if (ch[i] == '*') num++;
      if (ch[i] != '*' && ch[i - 1] == '*') {
        len[++cnt] = k * num + 1;
        num = 0;
      }
    }
    for (int i = cnt; i >= 1; --i) {
      ans[i] = x % len[i];
      x /= len[i];
    }
    num = 1;
    for (int i = 1; i <= n; ++i) {
      if (ch[i] == 'a')
        putchar('a');
      else {
        while (ch[i] == '*') i++;
        --i;
        for (int j = 1; j <= ans[num]; ++j) putchar('b');
        num++;
      }
    }
    putchar('\n');
  }
  return 0;
}

```