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.
2
55 982 31372
a*a*aa*aa*a*aa*aa*aaa*aaaaa*a*aa*aaaaa*aaaaa*aa*a*a*a*a
63 9 10946
*aaaaa*aa*a*a*aa*aaaa*a*aa*a*a*aaaa*aaaaa*a*a*aaa*aa*aaa*a*a*aa

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbba
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabaaaaabbbbbbbbbabbbbabbbbbaa

#include <bits/stdc++.h>
using namespace std;
int t, n, k, p[2010];
long long x, b[2010], ned[2010];
char s[2010];
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d%lld", &n, &k, &x);
    scanf("%s", s + 1);
    long long now = 0;
    int cnt = 0;
    for (int i = n; i >= 1; i--) cnt += (s[i] == 'a');
    if (x == 1) {
      for (int i = 1; i <= cnt; i++) putchar('a');
      putchar('\n');
      continue;
    }
    int lst = 0, tot = 0;
    for (int i = 1; i <= n; i++) {
      if (s[i] == 'a' && lst != i - 1) {
        b[++tot] = i - lst - 1, p[tot] = lst;
        lst = i;
      }
      if (s[i] == 'a') lst = i;
    }
    if (lst != n) b[++tot] = n - lst, p[tot] = lst;
    now = 1;
    for (int i = tot; i >= 1; i--) {
      now *= (b[i] * 1ll * k + 1ll);
      if (now >= x) {
        now /= (b[i] * 1ll * k + 1ll);
        for (int j = 1; j <= p[i]; j++)
          if (s[j] == 'a') putchar('a');
        if (now > 1) {
          int jn;
          for (jn = 1; now * 1ll * jn + now <= x; jn++) putchar('b');
          long long res = x - (now * 1ll * (jn - 1));
          memset(ned, 0, sizeof(ned));
          if (res) {
            if (res > now) putchar('b'), res -= now;
            int lst = n + 1;
            for (int j = n; j >= p[i] + b[i] + 1; j--) {
              if (s[j] == 'a') {
                ned[j] = min(res - 1, 1ll * k * (lst - j - 1));
                res -= k * (lst - j - 1) + 1;
                lst = j;
              }
            }
          }
        } else
          for (int j = 1; j <= x - now; j++) putchar('b');
        for (int j = p[i] + b[i] + 1; j <= n; j++) {
          if (s[j] == 'a') {
            putchar('a');
            for (int l = 1; l <= ned[j]; l++) putchar('b');
          }
        }
        putchar('\n');
        break;
      }
    }
  }
  return 0;
}
