#include <bits/stdc++.h>
using namespace std;
const long long int N = 20005;
template <typename T, typename T1>
T amax(T &a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T &a, T1 b) {
  if (b < a) a = b;
  return a;
}
long long int solve() {
  long long int n, k, x;
  cin >> n >> k >> x;
  string s;
  cin >> s;
  vector<long long int> p;
  long long int cur = 0;
  string nev;
  for (long long int i = 0; i < n; ++i) {
    if (s[i] == 'a') {
      if (cur) nev.push_back('*');
      nev.push_back('a');
      if (cur) p.push_back(cur);
      cur = 0;
    } else {
      cur++;
    }
  }
  if (cur) nev.push_back('*');
  if (cur) p.push_back(cur);
  long long int sz = p.size();
  vector<long long int> ans(sz, 0);
  long long int pw = 1;
  long long int idx = sz - 1;
  x--;
  while (x) {
    cur = k * p[idx] + 1;
    ans[idx] = x % cur;
    x /= cur;
    idx--;
  }
  string y;
  long long int j = 0;
  for (long long int i = 0; i < nev.size(); ++i) {
    if (nev[i] == 'a')
      y.push_back('a');
    else {
      long long int m = ans[j];
      for (long long int l = 0; l < m; ++l) y.push_back('b');
      j++;
    }
  }
  return cout << y << "\n", 0;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int test;
  cin >> test;
  while (test--) solve();
  return 0;
}

### Prompt

Please formulate 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 int N = 20005;
template <typename T, typename T1>
T amax(T &a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T &a, T1 b) {
  if (b < a) a = b;
  return a;
}
long long int solve() {
  long long int n, k, x;
  cin >> n >> k >> x;
  string s;
  cin >> s;
  vector<long long int> p;
  long long int cur = 0;
  string nev;
  for (long long int i = 0; i < n; ++i) {
    if (s[i] == 'a') {
      if (cur) nev.push_back('*');
      nev.push_back('a');
      if (cur) p.push_back(cur);
      cur = 0;
    } else {
      cur++;
    }
  }
  if (cur) nev.push_back('*');
  if (cur) p.push_back(cur);
  long long int sz = p.size();
  vector<long long int> ans(sz, 0);
  long long int pw = 1;
  long long int idx = sz - 1;
  x--;
  while (x) {
    cur = k * p[idx] + 1;
    ans[idx] = x % cur;
    x /= cur;
    idx--;
  }
  string y;
  long long int j = 0;
  for (long long int i = 0; i < nev.size(); ++i) {
    if (nev[i] == 'a')
      y.push_back('a');
    else {
      long long int m = ans[j];
      for (long long int l = 0; l < m; ++l) y.push_back('b');
      j++;
    }
  }
  return cout << y << "\n", 0;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int test;
  cin >> test;
  while (test--) solve();
  return 0;
}

```