#include <bits/stdc++.h>
using namespace std;
inline void debugMode() {}
void solvecases(int cse) {
  int n, k;
  long long x;
  cin >> n >> k >> x;
  string s;
  cin >> s;
  vector<long long> d;
  int cnt = 0;
  for (int i = 0; i < n; ++i) {
    if (s[i] == '*')
      ++cnt;
    else {
      if (cnt) {
        d.push_back(cnt);
        cnt = 0;
      }
    }
  }
  if (cnt) d.push_back(cnt);
  if (d.empty()) {
    cout << s << '\n';
    return;
  }
  if (k == 0) {
    for (auto e : s) {
      if (e == 'a') cout << 'a';
    }
    cout << '\n';
    return;
  }
  for (auto &e : d) {
    e = e * k + 1;
  }
  vector<long long> ans((int)d.size(), 0);
  vector<long long> m((int)d.size() + 1);
  for (int i = 0; i < (int)d.size(); ++i) m[i] = LLONG_MAX;
  m[(int)d.size()] = 1;
  for (int i = (int)d.size() - 1; i >= 0; --i) {
    if (log10((long double)d[i]) + log10((long double)m[i + 1]) > 18.0) break;
    m[i] = m[i + 1] * d[i];
  }
  --x;
  for (int i = 1; i < (int)d.size(); ++i) {
    if (m[i] != LLONG_MAX && m[i] <= x) {
      ans[i - 1] = x / m[i];
      x %= m[i];
    }
  }
  ans[(int)d.size() - 1] = x;
  int i = 0, j = 0;
  string b;
  while (i < n) {
    if (s[i] == '*') {
      for (int kk = 0; kk < ans[j]; ++kk) b += 'b';
      ++j;
      while (i < n && s[i] == '*') {
        ++i;
      }
    } else {
      b += 'a';
      ++i;
    }
  }
  cout << b << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    solvecases(i);
  }
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 debugMode() {}
void solvecases(int cse) {
  int n, k;
  long long x;
  cin >> n >> k >> x;
  string s;
  cin >> s;
  vector<long long> d;
  int cnt = 0;
  for (int i = 0; i < n; ++i) {
    if (s[i] == '*')
      ++cnt;
    else {
      if (cnt) {
        d.push_back(cnt);
        cnt = 0;
      }
    }
  }
  if (cnt) d.push_back(cnt);
  if (d.empty()) {
    cout << s << '\n';
    return;
  }
  if (k == 0) {
    for (auto e : s) {
      if (e == 'a') cout << 'a';
    }
    cout << '\n';
    return;
  }
  for (auto &e : d) {
    e = e * k + 1;
  }
  vector<long long> ans((int)d.size(), 0);
  vector<long long> m((int)d.size() + 1);
  for (int i = 0; i < (int)d.size(); ++i) m[i] = LLONG_MAX;
  m[(int)d.size()] = 1;
  for (int i = (int)d.size() - 1; i >= 0; --i) {
    if (log10((long double)d[i]) + log10((long double)m[i + 1]) > 18.0) break;
    m[i] = m[i + 1] * d[i];
  }
  --x;
  for (int i = 1; i < (int)d.size(); ++i) {
    if (m[i] != LLONG_MAX && m[i] <= x) {
      ans[i - 1] = x / m[i];
      x %= m[i];
    }
  }
  ans[(int)d.size() - 1] = x;
  int i = 0, j = 0;
  string b;
  while (i < n) {
    if (s[i] == '*') {
      for (int kk = 0; kk < ans[j]; ++kk) b += 'b';
      ++j;
      while (i < n && s[i] == '*') {
        ++i;
      }
    } else {
      b += 'a';
      ++i;
    }
  }
  cout << b << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    solvecases(i);
  }
}

```