#include <bits/stdc++.h>
using namespace std;
template <typename T>
ostream& operator<<(ostream& os, vector<T>& v) {
  if (v.size() == 0) {
    os << "empty vector\n";
    return os;
  }
  for (auto element : v) os << element << " ";
  return os;
}
template <typename T, typename second>
ostream& operator<<(ostream& os, pair<T, second>& p) {
  os << "(" << p.first << ", " << p.second << ")";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, set<T>& v) {
  if (v.size() == 0) {
    os << "empty set\n";
    return os;
  }
  auto endit = v.end();
  endit--;
  os << "[";
  for (auto it = v.begin(); it != v.end(); it++) {
    os << *it;
    if (it != endit) os << ", ";
  }
  os << "]";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, multiset<T>& v) {
  if (v.size() == 0) {
    os << "empty multiset\n";
    return os;
  }
  auto endit = v.end();
  endit--;
  os << "[";
  for (auto it = v.begin(); it != v.end(); it++) {
    os << *it;
    if (it != endit) os << ", ";
  }
  os << "]";
  return os;
}
template <typename T, typename second>
ostream& operator<<(ostream& os, map<T, second>& v) {
  if (v.size() == 0) {
    os << "empty map\n";
    return os;
  }
  auto endit = v.end();
  endit--;
  os << "{";
  for (auto it = v.begin(); it != v.end(); it++) {
    os << "(" << (*it).first << " : " << (*it).second << ")";
    if (it != endit) os << ", ";
  }
  os << "}";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, vector<vector<T>>& v) {
  for (auto& subv : v) {
    for (auto& e : subv) os << e << " ";
    os << "\n";
  }
  return os;
}
bool do_debug = false;
void Runtime_Terror() {
  long long n, k, tem;
  cin >> n >> k >> tem;
  long long x = tem;
  vector<string> v;
  string s;
  cin >> s;
  vector<long long> pos, maxi;
  for (long long i = 0; i < s.size(); i++) {
    if (s[i] == '*') {
      long long cnt = 0;
      pos.push_back(i);
      while (i < s.size() && s[i] == '*') {
        cnt++;
        i++;
      }
      maxi.push_back(cnt * k + 1);
    }
  }
  if (maxi.size() == 0) {
    cout << s << "\n";
    return;
  };
  ;
  vector<long long> ans(pos.size());
  vector<long long> pro(maxi.size());
  pro[pro.size() - 1] = maxi.back();
  long long sz = maxi.size();
  for (long long i = sz - 2; i >= 0; i--) {
    if (pro[i + 1] == -1) {
      pro[i] = -1;
      continue;
    }
    long double cal = (LLONG_MAX - 1e8) / (maxi[i] * 1.0);
    if (pro[i + 1] > (long long)(cal)) {
      pro[i] = -1;
      continue;
    }
    pro[i] = pro[i + 1] * maxi[i];
  }
  for (long long i = 0; i < maxi.size() - 1; i++) {
    if (pro[i + 1] == -1) {
      ans[i] = 1;
      continue;
    }
    long long small = pro[i + 1];
    for (long long j = 0; j < maxi[i] + 1; j++) {
      if (small >= x) {
        ans[i] = j + 1;
        break;
      }
      small += pro[i + 1];
    }
    x -= (ans[i] - 1) * (pro[i + 1]);
  }
  ans[ans.size() - 1] = x;
  ;
  long long in = 0;
  for (long long i = 0; i < s.size(); i++) {
    if (s[i] != '*')
      cout << s[i];
    else {
      for (long long j = 0; j < ans[in] - 1; j++) cout << "b";
      in++;
      while (i < s.size() && s[i] == '*') i++;
      i--;
    }
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  for (long long i = 0; i < t; i++) Runtime_Terror();
  return 0;
}

### Prompt

Your task is to create 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>
ostream& operator<<(ostream& os, vector<T>& v) {
  if (v.size() == 0) {
    os << "empty vector\n";
    return os;
  }
  for (auto element : v) os << element << " ";
  return os;
}
template <typename T, typename second>
ostream& operator<<(ostream& os, pair<T, second>& p) {
  os << "(" << p.first << ", " << p.second << ")";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, set<T>& v) {
  if (v.size() == 0) {
    os << "empty set\n";
    return os;
  }
  auto endit = v.end();
  endit--;
  os << "[";
  for (auto it = v.begin(); it != v.end(); it++) {
    os << *it;
    if (it != endit) os << ", ";
  }
  os << "]";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, multiset<T>& v) {
  if (v.size() == 0) {
    os << "empty multiset\n";
    return os;
  }
  auto endit = v.end();
  endit--;
  os << "[";
  for (auto it = v.begin(); it != v.end(); it++) {
    os << *it;
    if (it != endit) os << ", ";
  }
  os << "]";
  return os;
}
template <typename T, typename second>
ostream& operator<<(ostream& os, map<T, second>& v) {
  if (v.size() == 0) {
    os << "empty map\n";
    return os;
  }
  auto endit = v.end();
  endit--;
  os << "{";
  for (auto it = v.begin(); it != v.end(); it++) {
    os << "(" << (*it).first << " : " << (*it).second << ")";
    if (it != endit) os << ", ";
  }
  os << "}";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, vector<vector<T>>& v) {
  for (auto& subv : v) {
    for (auto& e : subv) os << e << " ";
    os << "\n";
  }
  return os;
}
bool do_debug = false;
void Runtime_Terror() {
  long long n, k, tem;
  cin >> n >> k >> tem;
  long long x = tem;
  vector<string> v;
  string s;
  cin >> s;
  vector<long long> pos, maxi;
  for (long long i = 0; i < s.size(); i++) {
    if (s[i] == '*') {
      long long cnt = 0;
      pos.push_back(i);
      while (i < s.size() && s[i] == '*') {
        cnt++;
        i++;
      }
      maxi.push_back(cnt * k + 1);
    }
  }
  if (maxi.size() == 0) {
    cout << s << "\n";
    return;
  };
  ;
  vector<long long> ans(pos.size());
  vector<long long> pro(maxi.size());
  pro[pro.size() - 1] = maxi.back();
  long long sz = maxi.size();
  for (long long i = sz - 2; i >= 0; i--) {
    if (pro[i + 1] == -1) {
      pro[i] = -1;
      continue;
    }
    long double cal = (LLONG_MAX - 1e8) / (maxi[i] * 1.0);
    if (pro[i + 1] > (long long)(cal)) {
      pro[i] = -1;
      continue;
    }
    pro[i] = pro[i + 1] * maxi[i];
  }
  for (long long i = 0; i < maxi.size() - 1; i++) {
    if (pro[i + 1] == -1) {
      ans[i] = 1;
      continue;
    }
    long long small = pro[i + 1];
    for (long long j = 0; j < maxi[i] + 1; j++) {
      if (small >= x) {
        ans[i] = j + 1;
        break;
      }
      small += pro[i + 1];
    }
    x -= (ans[i] - 1) * (pro[i + 1]);
  }
  ans[ans.size() - 1] = x;
  ;
  long long in = 0;
  for (long long i = 0; i < s.size(); i++) {
    if (s[i] != '*')
      cout << s[i];
    else {
      for (long long j = 0; j < ans[in] - 1; j++) cout << "b";
      in++;
      while (i < s.size() && s[i] == '*') i++;
      i--;
    }
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  for (long long i = 0; i < t; i++) Runtime_Terror();
  return 0;
}

```