#include <bits/stdc++.h>
using namespace std;
inline void read(long long &x) {
  x = 0;
  long long f = 1;
  char ch = getchar();
  while (ch != 45 && (ch > '9' || ch < '0')) ch = getchar();
  if (ch == 45) {
    f = -1, ch = getchar();
  }
  while (ch <= '9' && ch >= '0') {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  x *= f;
}
void cmin(long long &a, long long b) { a = min(a, b); }
void cmax(long long &a, long long b) { a = max(a, b); }
const long long N = 2e3 + 10;
long long i, j, k, n, s, t, m;
char ch[N], ans[N];
long long a[N], b[N];
signed main() {
  read(t);
  while (t--) {
    read(n), read(k), read(s);
    m = 0;
    scanf("%s", ch + 1);
    memset(a, 0, sizeof(a));
    for (i = 1; i <= n; i++) {
      if (ch[i] != ch[i - 1]) ++m;
      if (ch[i] == 'a')
        a[m]--;
      else
        a[m]++;
    }
    s--;
    for (i = m; i >= 1; i--) {
      if (a[i] < 0) {
        b[i] = a[i];
        continue;
      }
      b[i] = s % (a[i] * k + 1);
      s /= (a[i] * k + 1);
    }
    for (i = 1; i <= m; i++) {
      for (j = 1; j <= abs(b[i]); j++) putchar((b[i] < 0 ? 'a' : 'b'));
    }
    cout << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
inline void read(long long &x) {
  x = 0;
  long long f = 1;
  char ch = getchar();
  while (ch != 45 && (ch > '9' || ch < '0')) ch = getchar();
  if (ch == 45) {
    f = -1, ch = getchar();
  }
  while (ch <= '9' && ch >= '0') {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  x *= f;
}
void cmin(long long &a, long long b) { a = min(a, b); }
void cmax(long long &a, long long b) { a = max(a, b); }
const long long N = 2e3 + 10;
long long i, j, k, n, s, t, m;
char ch[N], ans[N];
long long a[N], b[N];
signed main() {
  read(t);
  while (t--) {
    read(n), read(k), read(s);
    m = 0;
    scanf("%s", ch + 1);
    memset(a, 0, sizeof(a));
    for (i = 1; i <= n; i++) {
      if (ch[i] != ch[i - 1]) ++m;
      if (ch[i] == 'a')
        a[m]--;
      else
        a[m]++;
    }
    s--;
    for (i = m; i >= 1; i--) {
      if (a[i] < 0) {
        b[i] = a[i];
        continue;
      }
      b[i] = s % (a[i] * k + 1);
      s /= (a[i] * k + 1);
    }
    for (i = 1; i <= m; i++) {
      for (j = 1; j <= abs(b[i]); j++) putchar((b[i] < 0 ? 'a' : 'b'));
    }
    cout << endl;
  }
  return 0;
}

```