#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
static inline void amn(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static inline void amx(T &x, U y) {
  if (x < y) x = y;
}
const long long MOD = 1e9 + 7;
const long long INF = 2e18;
const long long N = 5e5 + 5;
long long n, k, x;
string s, t;
void solve() {
  cin >> n >> k >> x;
  cin >> s;
  t = "";
  vector<long long> v, vec;
  for (long long i = 0; i < n; i++) {
    t.push_back(s[i]);
    if (s[i] == '*') {
      long long j = i;
      while (j < n and s[j] == '*') {
        j++;
      }
      v.push_back(j - i);
      vec.push_back(0);
      i = j - 1;
    }
  }
  long long m = v.size();
  vector<long long> product(m + 1, 1);
  if (m > 0) product[m - 1] = v[m - 1] * k + 1;
  for (long long i = m - 2; i >= 0; i--) {
    long long p = (INF / product[i + 1]) + 1;
    if ((v[i] * k + 1) > p) {
      product[i] = INF;
    } else {
      product[i] = product[i + 1] * (v[i] * k + 1);
    }
    amn(product[i], INF);
  }
  for (long long i = 0; i < m; i++) {
    if (product[i + 1] >= x) {
      continue;
    }
    long long tmp = product[i + 1];
    vec[i] = (x - 1) / tmp;
    x -= vec[i] * tmp;
  }
  n = t.size();
  string ans;
  long long c = 0;
  for (long long i = 0; i < n; i++) {
    if (t[i] == 'a') {
      ans.push_back('a');
    } else {
      for (long long j = 1; j <= vec[c]; j++) {
        ans.push_back('b');
      }
      c++;
    }
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long test_case = 1;
  cin >> test_case;
  while (test_case--) {
    solve();
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
template <typename T, typename U>
static inline void amn(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static inline void amx(T &x, U y) {
  if (x < y) x = y;
}
const long long MOD = 1e9 + 7;
const long long INF = 2e18;
const long long N = 5e5 + 5;
long long n, k, x;
string s, t;
void solve() {
  cin >> n >> k >> x;
  cin >> s;
  t = "";
  vector<long long> v, vec;
  for (long long i = 0; i < n; i++) {
    t.push_back(s[i]);
    if (s[i] == '*') {
      long long j = i;
      while (j < n and s[j] == '*') {
        j++;
      }
      v.push_back(j - i);
      vec.push_back(0);
      i = j - 1;
    }
  }
  long long m = v.size();
  vector<long long> product(m + 1, 1);
  if (m > 0) product[m - 1] = v[m - 1] * k + 1;
  for (long long i = m - 2; i >= 0; i--) {
    long long p = (INF / product[i + 1]) + 1;
    if ((v[i] * k + 1) > p) {
      product[i] = INF;
    } else {
      product[i] = product[i + 1] * (v[i] * k + 1);
    }
    amn(product[i], INF);
  }
  for (long long i = 0; i < m; i++) {
    if (product[i + 1] >= x) {
      continue;
    }
    long long tmp = product[i + 1];
    vec[i] = (x - 1) / tmp;
    x -= vec[i] * tmp;
  }
  n = t.size();
  string ans;
  long long c = 0;
  for (long long i = 0; i < n; i++) {
    if (t[i] == 'a') {
      ans.push_back('a');
    } else {
      for (long long j = 1; j <= vec[c]; j++) {
        ans.push_back('b');
      }
      c++;
    }
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long test_case = 1;
  cin >> test_case;
  while (test_case--) {
    solve();
  }
  return 0;
}

```