#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
const long long N = 1e5 + 10;
template <typename... T>
void debug(T&... args) {
  cout << "\n==================\n";
  ((cout << args << ", "), ...);
  cout << "\n==================\n";
}
template <typename T>
void debug(vector<T>& a) {
  cout << "\n==============\n";
  for (auto& e : a) cout << e << ' ';
  cout << "\n==============\n";
}
template <typename... T>
void put(T&... args) {
  ((cout << args << " "), ...);
}
void solve() {
  long long n, k, x;
  string s;
  cin >> n >> k >> x >> s;
  vector<long long> a;
  for (long long i = 0; i < n;) {
    if (s[i] == 'a') {
      a.push_back(-1);
      i++;
    } else {
      long long c = 0;
      while (i < n && s[i] == '*') i++, c += k;
      a.push_back(c);
    }
  }
  vector<long long> b(a.size() + 1);
  b[a.size()] = 1;
  x--;
  for (long long i = a.size() - 1; i > 0; i--) {
    long long prv = b[i + 1];
    if (a[i] < 0)
      b[i] = prv;
    else {
      if (a[i] + 1 > x / prv) {
        b[i] = x + 1;
      } else
        b[i] = prv * (a[i] + 1);
    }
  }
  string ans = "";
  for (long long i = 0; i < a.size(); i++) {
    if (a[i] < 0)
      ans += 'a';
    else {
      long long nxt = b[i + 1];
      while (x >= nxt) x -= nxt, ans += 'b';
    }
  }
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Your challenge is to write 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 mod = 998244353;
const long long N = 1e5 + 10;
template <typename... T>
void debug(T&... args) {
  cout << "\n==================\n";
  ((cout << args << ", "), ...);
  cout << "\n==================\n";
}
template <typename T>
void debug(vector<T>& a) {
  cout << "\n==============\n";
  for (auto& e : a) cout << e << ' ';
  cout << "\n==============\n";
}
template <typename... T>
void put(T&... args) {
  ((cout << args << " "), ...);
}
void solve() {
  long long n, k, x;
  string s;
  cin >> n >> k >> x >> s;
  vector<long long> a;
  for (long long i = 0; i < n;) {
    if (s[i] == 'a') {
      a.push_back(-1);
      i++;
    } else {
      long long c = 0;
      while (i < n && s[i] == '*') i++, c += k;
      a.push_back(c);
    }
  }
  vector<long long> b(a.size() + 1);
  b[a.size()] = 1;
  x--;
  for (long long i = a.size() - 1; i > 0; i--) {
    long long prv = b[i + 1];
    if (a[i] < 0)
      b[i] = prv;
    else {
      if (a[i] + 1 > x / prv) {
        b[i] = x + 1;
      } else
        b[i] = prv * (a[i] + 1);
    }
  }
  string ans = "";
  for (long long i = 0; i < a.size(); i++) {
    if (a[i] < 0)
      ans += 'a';
    else {
      long long nxt = b[i + 1];
      while (x >= nxt) x -= nxt, ans += 'b';
    }
  }
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) solve();
}

```