#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
template <class T>
using pq = priority_queue<T>;
template <class T>
using pqg = priority_queue<T, vector<T>, greater<T>>;
template <class T>
bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long MOD = 1000000007;
const char nl = '\n';
const long long MX = 200001;
void solve() {
  long long n, k, x;
  cin >> n >> k >> x;
  string s;
  cin >> s;
  vector<long long> v;
  long long cnt = 0;
  for (long long i = 0; i < s.size(); ++i) {
    if (s[i] != '*') {
      if (cnt) v.push_back(cnt * k);
      cnt = 0;
      continue;
    } else
      cnt++;
  }
  if (cnt) v.push_back(cnt * k);
  vector<long long> ans(v.size(), 0);
  while (!v.empty() && x != 0) {
    if (x <= v.back() + 1) {
      if (!v.empty()) ans[v.size() - 1] = min(x - 1, 1ll * v.back());
      break;
    }
    long long cur = v.back() + 1;
    for (long long i = v.size() - 2; i >= 0; --i) {
      if ((x - cur) / (v[i]) <= cur) {
        long long d = (x - cur) / (cur) + (((x - cur) % (cur)) > 0);
        ans[i] = d;
        x = x - (d * (cur));
        break;
      }
      cur += (v[i]) * (cur);
    }
  }
  long long idx = 0;
  for (long long i = 0; i < s.size(); ++i) {
    if (s[i] != '*') {
      if (i && s[i - 1] == '*') {
        while (ans[idx]--) cout << 'b';
        idx++;
      }
      cout << s[i];
    }
  }
  if (idx < v.size()) {
    while (ans[idx]--) cout << 'b';
    idx++;
  }
  cout << nl;
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(cin.failbit);
  long long T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}

### 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>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
template <class T>
using pq = priority_queue<T>;
template <class T>
using pqg = priority_queue<T, vector<T>, greater<T>>;
template <class T>
bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long MOD = 1000000007;
const char nl = '\n';
const long long MX = 200001;
void solve() {
  long long n, k, x;
  cin >> n >> k >> x;
  string s;
  cin >> s;
  vector<long long> v;
  long long cnt = 0;
  for (long long i = 0; i < s.size(); ++i) {
    if (s[i] != '*') {
      if (cnt) v.push_back(cnt * k);
      cnt = 0;
      continue;
    } else
      cnt++;
  }
  if (cnt) v.push_back(cnt * k);
  vector<long long> ans(v.size(), 0);
  while (!v.empty() && x != 0) {
    if (x <= v.back() + 1) {
      if (!v.empty()) ans[v.size() - 1] = min(x - 1, 1ll * v.back());
      break;
    }
    long long cur = v.back() + 1;
    for (long long i = v.size() - 2; i >= 0; --i) {
      if ((x - cur) / (v[i]) <= cur) {
        long long d = (x - cur) / (cur) + (((x - cur) % (cur)) > 0);
        ans[i] = d;
        x = x - (d * (cur));
        break;
      }
      cur += (v[i]) * (cur);
    }
  }
  long long idx = 0;
  for (long long i = 0; i < s.size(); ++i) {
    if (s[i] != '*') {
      if (i && s[i - 1] == '*') {
        while (ans[idx]--) cout << 'b';
        idx++;
      }
      cout << s[i];
    }
  }
  if (idx < v.size()) {
    while (ans[idx]--) cout << 'b';
    idx++;
  }
  cout << nl;
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(cin.failbit);
  long long T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}

```