#include <bits/stdc++.h>
using namespace std;
template <typename T, typename T_iterable>
vector<pair<T, int>> run_length_encoding(const T_iterable &items) {
  vector<pair<T, int>> encoding;
  T previous;
  int count = 0;
  for (const T &item : items)
    if (item == previous) {
      count++;
    } else {
      if (count > 0) encoding.emplace_back(previous, count);
      previous = item;
      count = 1;
    }
  if (count > 0) encoding.emplace_back(previous, count);
  return encoding;
}
void run_case() {
  int N, K;
  int64_t X;
  string S;
  cin >> N >> K >> X >> S;
  X--;
  vector<pair<char, int>> encoding = run_length_encoding<char>(S);
  vector<int64_t> bases;
  for (auto &pr : encoding)
    if (pr.first == '*') bases.push_back(pr.second * K + 1);
  vector<int64_t> B;
  while (!bases.empty()) {
    B.push_back(X % bases.back());
    X /= bases.back();
    bases.pop_back();
  }
  string ans;
  for (auto &pr : encoding)
    if (pr.first == '*') {
      ans += string(B.back(), 'b');
      B.pop_back();
    } else {
      ans += string(pr.second, pr.first);
    }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int tests;
  cin >> tests;
  while (tests-- > 0) run_case();
}

### 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;
template <typename T, typename T_iterable>
vector<pair<T, int>> run_length_encoding(const T_iterable &items) {
  vector<pair<T, int>> encoding;
  T previous;
  int count = 0;
  for (const T &item : items)
    if (item == previous) {
      count++;
    } else {
      if (count > 0) encoding.emplace_back(previous, count);
      previous = item;
      count = 1;
    }
  if (count > 0) encoding.emplace_back(previous, count);
  return encoding;
}
void run_case() {
  int N, K;
  int64_t X;
  string S;
  cin >> N >> K >> X >> S;
  X--;
  vector<pair<char, int>> encoding = run_length_encoding<char>(S);
  vector<int64_t> bases;
  for (auto &pr : encoding)
    if (pr.first == '*') bases.push_back(pr.second * K + 1);
  vector<int64_t> B;
  while (!bases.empty()) {
    B.push_back(X % bases.back());
    X /= bases.back();
    bases.pop_back();
  }
  string ans;
  for (auto &pr : encoding)
    if (pr.first == '*') {
      ans += string(B.back(), 'b');
      B.pop_back();
    } else {
      ans += string(pr.second, pr.first);
    }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int tests;
  cin >> tests;
  while (tests-- > 0) run_case();
}

```