#include <bits/stdc++.h>
using namespace std;
int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
int dx[] = {0, 0, 1, -1, 1, -1, 1, -1};
void file() {}
void fast() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
}
long long mul(long long a, long long b) {
  if (a <= LLONG_MAX / b) return a * b;
  return (long long)1e18;
}
void test_case() {
  int n, k;
  long long x;
  cin >> n >> k >> x;
  string str;
  cin >> str;
  string tmp = "";
  vector<int> can;
  for (auto& ch : str) {
    if (ch == 'a') {
      can.push_back(0);
      tmp.push_back(ch);
      continue;
    }
    if (tmp.empty() || tmp.back() == 'a')
      tmp.push_back(ch), can.push_back(k);
    else
      can.back() += k;
  }
  str = tmp;
  tmp = "";
  n = (int)(str.size());
  vector<long long> suf(n + 1);
  suf.back() = 1;
  for (int i = n - 1; i >= 0; i--)
    suf[i] = (str[i] == 'a' ? suf[i + 1] : mul(can[i] + 1, suf[i + 1]));
  for (int i = 0; i < n; i++) {
    if (str[i] == 'a') {
      tmp.push_back('a');
      continue;
    }
    for (int j = 0; j <= can[i]; j++) {
      if (suf[i + 1] >= x) break;
      x -= suf[i + 1];
      tmp.push_back('b');
    }
  }
  cout << tmp << '\n';
}
int main() {
  file();
  fast();
  int t;
  cin >> t;
  while (t--) {
    test_case();
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
int dx[] = {0, 0, 1, -1, 1, -1, 1, -1};
void file() {}
void fast() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
}
long long mul(long long a, long long b) {
  if (a <= LLONG_MAX / b) return a * b;
  return (long long)1e18;
}
void test_case() {
  int n, k;
  long long x;
  cin >> n >> k >> x;
  string str;
  cin >> str;
  string tmp = "";
  vector<int> can;
  for (auto& ch : str) {
    if (ch == 'a') {
      can.push_back(0);
      tmp.push_back(ch);
      continue;
    }
    if (tmp.empty() || tmp.back() == 'a')
      tmp.push_back(ch), can.push_back(k);
    else
      can.back() += k;
  }
  str = tmp;
  tmp = "";
  n = (int)(str.size());
  vector<long long> suf(n + 1);
  suf.back() = 1;
  for (int i = n - 1; i >= 0; i--)
    suf[i] = (str[i] == 'a' ? suf[i + 1] : mul(can[i] + 1, suf[i + 1]));
  for (int i = 0; i < n; i++) {
    if (str[i] == 'a') {
      tmp.push_back('a');
      continue;
    }
    for (int j = 0; j <= can[i]; j++) {
      if (suf[i + 1] >= x) break;
      x -= suf[i + 1];
      tmp.push_back('b');
    }
  }
  cout << tmp << '\n';
}
int main() {
  file();
  fast();
  int t;
  cin >> t;
  while (t--) {
    test_case();
  }
  return 0;
}

```