#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 7, inf = 2e9 + 9, mod = 1e9 + 7;
const long long linf = 2e18 + 7;
int n, k;
long long x, d[maxn];
void solve() {
  vector<long long> a;
  char s[maxn], dum[maxn];
  scanf("%d%d%I64d", &n, &k, &x);
  memset(d, 0, (n + 3) * sizeof(long long));
  scanf("%s", &dum);
  int l = 0;
  for (int i = 0; i < n + 1; ++i) {
    if (dum[i] == '*')
      l++;
    else {
      a.push_back(1ll * l * k);
      l = 0;
    }
  }
  int sz = a.size();
  a.push_back(0);
  long long t = 1;
  for (int i = sz - 1; i >= 0; --i) {
    if (linf / t < (a[i] + 1) or t * (a[i] + 1) >= x) {
      x--;
      for (int j = i; j < sz; ++j) {
        d[j] = x / t;
        x -= d[j] * t;
        t /= a[j + 1] + 1;
      }
      break;
    }
    t = t * (a[i] + 1);
  }
  for (int i = 0; i < sz; ++i) {
    for (int j = 0; j < d[i]; ++j) {
      printf("b");
    }
    if (i != sz - 1) printf("a");
  }
  puts("");
}
int main() {
  int test;
  scanf("%d", &test);
  while (test--) solve();
  return 0;
}

### Prompt

Create a solution in cpp for 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 int maxn = 2e5 + 7, inf = 2e9 + 9, mod = 1e9 + 7;
const long long linf = 2e18 + 7;
int n, k;
long long x, d[maxn];
void solve() {
  vector<long long> a;
  char s[maxn], dum[maxn];
  scanf("%d%d%I64d", &n, &k, &x);
  memset(d, 0, (n + 3) * sizeof(long long));
  scanf("%s", &dum);
  int l = 0;
  for (int i = 0; i < n + 1; ++i) {
    if (dum[i] == '*')
      l++;
    else {
      a.push_back(1ll * l * k);
      l = 0;
    }
  }
  int sz = a.size();
  a.push_back(0);
  long long t = 1;
  for (int i = sz - 1; i >= 0; --i) {
    if (linf / t < (a[i] + 1) or t * (a[i] + 1) >= x) {
      x--;
      for (int j = i; j < sz; ++j) {
        d[j] = x / t;
        x -= d[j] * t;
        t /= a[j + 1] + 1;
      }
      break;
    }
    t = t * (a[i] + 1);
  }
  for (int i = 0; i < sz; ++i) {
    for (int j = 0; j < d[i]; ++j) {
      printf("b");
    }
    if (i != sz - 1) printf("a");
  }
  puts("");
}
int main() {
  int test;
  scanf("%d", &test);
  while (test--) solve();
  return 0;
}

```