#include <bits/stdc++.h>
using namespace std;
using ll = long long;
constexpr int INF = 0x3f3f3f3f;
constexpr long long LINF = 0x3f3f3f3f3f3f3f3fLL;
constexpr double EPS = 1e-8;
constexpr int MOD = 1000000007;
constexpr int DY[]{1, 0, -1, 0}, DX[]{0, -1, 0, 1};
constexpr int DY8[]{1, 1, 0, -1, -1, -1, 0, 1},
    DX8[]{0, -1, -1, -1, 0, 1, 1, 1};
template <typename T, typename U>
inline bool chmax(T& a, U b) {
  return a < b ? (a = b, true) : false;
}
template <typename T, typename U>
inline bool chmin(T& a, U b) {
  return a > b ? (a = b, true) : false;
}
struct IOSetup {
  IOSetup() {
    std::cin.tie(nullptr);
    std::ios_base::sync_with_stdio(false);
    std::cout << fixed << setprecision(20);
  }
} iosetup;
template <typename T = char, typename U = std::string>
std::vector<std::pair<T, int>> run_length_encoding(const U& s) {
  int n = s.size();
  std::vector<std::pair<T, int>> res;
  T now = s[0];
  int cnt = 1;
  for (int i = 1; i < n; ++i) {
    if (s[i] != now) {
      res.emplace_back(now, cnt);
      cnt = 0;
    }
    now = s[i];
    ++cnt;
  }
  res.emplace_back(now, cnt);
  return res;
}
void solve() {
  int n, k;
  ll x;
  cin >> n >> k >> x;
  string s;
  cin >> s;
  const auto rle = run_length_encoding(s);
  const int m = rle.size();
  string ans = "";
  vector<ll> w(m, 0);
  for (int i = m - 1; i >= 0; --i) {
    if (rle[i].first == '*') {
      const ll num = 1LL * k * rle[i].second + 1,
               r = (i + 2 < m ? w[i + 2] : 1);
      w[i] = (r > LINF / num ? LINF : r * num);
    }
  }
  for (int i = (0); i < (m); ++i) {
    if (rle[i].first == 'a') {
      ans += string(rle[i].second, 'a');
      continue;
    }
    const ll r = (i + 2 < m ? w[i + 2] : 1);
    const ll put = (x - 1) / r;
    x -= put * r;
    ans += string(put, 'b');
  }
  assert(x == 1);
  cout << ans << '\n';
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Develop a solution in CPP 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;
using ll = long long;
constexpr int INF = 0x3f3f3f3f;
constexpr long long LINF = 0x3f3f3f3f3f3f3f3fLL;
constexpr double EPS = 1e-8;
constexpr int MOD = 1000000007;
constexpr int DY[]{1, 0, -1, 0}, DX[]{0, -1, 0, 1};
constexpr int DY8[]{1, 1, 0, -1, -1, -1, 0, 1},
    DX8[]{0, -1, -1, -1, 0, 1, 1, 1};
template <typename T, typename U>
inline bool chmax(T& a, U b) {
  return a < b ? (a = b, true) : false;
}
template <typename T, typename U>
inline bool chmin(T& a, U b) {
  return a > b ? (a = b, true) : false;
}
struct IOSetup {
  IOSetup() {
    std::cin.tie(nullptr);
    std::ios_base::sync_with_stdio(false);
    std::cout << fixed << setprecision(20);
  }
} iosetup;
template <typename T = char, typename U = std::string>
std::vector<std::pair<T, int>> run_length_encoding(const U& s) {
  int n = s.size();
  std::vector<std::pair<T, int>> res;
  T now = s[0];
  int cnt = 1;
  for (int i = 1; i < n; ++i) {
    if (s[i] != now) {
      res.emplace_back(now, cnt);
      cnt = 0;
    }
    now = s[i];
    ++cnt;
  }
  res.emplace_back(now, cnt);
  return res;
}
void solve() {
  int n, k;
  ll x;
  cin >> n >> k >> x;
  string s;
  cin >> s;
  const auto rle = run_length_encoding(s);
  const int m = rle.size();
  string ans = "";
  vector<ll> w(m, 0);
  for (int i = m - 1; i >= 0; --i) {
    if (rle[i].first == '*') {
      const ll num = 1LL * k * rle[i].second + 1,
               r = (i + 2 < m ? w[i + 2] : 1);
      w[i] = (r > LINF / num ? LINF : r * num);
    }
  }
  for (int i = (0); i < (m); ++i) {
    if (rle[i].first == 'a') {
      ans += string(rle[i].second, 'a');
      continue;
    }
    const ll r = (i + 2 < m ? w[i + 2] : 1);
    const ll put = (x - 1) / r;
    x -= put * r;
    ans += string(put, 'b');
  }
  assert(x == 1);
  cout << ans << '\n';
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```