#include <bits/stdc++.h>
using namespace std;
const long long N = 2e3 + 7;
long long n;
char a[N];
long long k;
long long x;
long long mim[N];
long long ban[N];
long long s[N];
long long xx = 0;
long long gon;
void solve() {
  long long cnt = 0;
  string res = "";
  cin >> n >> k >> x;
  long long dak = 1;
  xx++;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    ban[i] = 0;
    mim[i] = 0;
  }
  for (long long i = 1; i <= n; i++) {
    if (a[i] == '*') {
      long long dem = 0;
      long long j = i;
      while (j <= n && a[j] == '*') {
        dem++;
        j++;
      }
      mim[++cnt] = (j - i) * k + 1;
      i = j - 1;
    }
  }
  long long pos = cnt + 1;
  for (long long i = cnt; i >= 1; i--) {
    if (dak >= x / mim[i] + bool(x % mim[i])) {
      pos = i;
      break;
    }
    dak = dak * mim[i];
    s[i] = dak;
  }
  s[cnt + 1] = 1;
  for (long long i = pos; i <= cnt; i++) {
    ban[i] = x / s[i + 1] + bool(x % s[i + 1]) - 1;
    x = x % s[i + 1];
    if (x == 0) {
      x = s[i + 1];
    }
  }
  long long id = 0;
  for (long long i = 1; i <= n; i++) {
    if (a[i] == 'a') {
      cout << 'a';
    } else {
      long long j = i;
      while (j <= n && a[j] == '*') {
        j++;
      }
      id++;
      for (long long t = 1; t <= ban[id]; t++) {
        cout << 'b';
      }
      i = j - 1;
    }
  }
  cout << '\n';
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long test;
  cin >> test;
  gon = test;
  while (test--) {
    solve();
  }
}

### Prompt

Generate 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;
const long long N = 2e3 + 7;
long long n;
char a[N];
long long k;
long long x;
long long mim[N];
long long ban[N];
long long s[N];
long long xx = 0;
long long gon;
void solve() {
  long long cnt = 0;
  string res = "";
  cin >> n >> k >> x;
  long long dak = 1;
  xx++;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    ban[i] = 0;
    mim[i] = 0;
  }
  for (long long i = 1; i <= n; i++) {
    if (a[i] == '*') {
      long long dem = 0;
      long long j = i;
      while (j <= n && a[j] == '*') {
        dem++;
        j++;
      }
      mim[++cnt] = (j - i) * k + 1;
      i = j - 1;
    }
  }
  long long pos = cnt + 1;
  for (long long i = cnt; i >= 1; i--) {
    if (dak >= x / mim[i] + bool(x % mim[i])) {
      pos = i;
      break;
    }
    dak = dak * mim[i];
    s[i] = dak;
  }
  s[cnt + 1] = 1;
  for (long long i = pos; i <= cnt; i++) {
    ban[i] = x / s[i + 1] + bool(x % s[i + 1]) - 1;
    x = x % s[i + 1];
    if (x == 0) {
      x = s[i + 1];
    }
  }
  long long id = 0;
  for (long long i = 1; i <= n; i++) {
    if (a[i] == 'a') {
      cout << 'a';
    } else {
      long long j = i;
      while (j <= n && a[j] == '*') {
        j++;
      }
      id++;
      for (long long t = 1; t <= ban[id]; t++) {
        cout << 'b';
      }
      i = j - 1;
    }
  }
  cout << '\n';
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long test;
  cin >> test;
  gon = test;
  while (test--) {
    solve();
  }
}

```