#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
const long long MAX = 3e5 + 10;
const long long Time = 86400;
const long long X = 131;
const long long inf = 0x3f3f3f3f;
const double PI = 1e-4;
double pai = 3.14159265358979323846;
long long T, n, m, k, x, ans;
string s;
long long a[MAX];
vector<long long> v;
char ch;
void solve() {
  v.clear();
  cin >> n >> k >> x;
  for (long long i = 1; i <= n; i++) {
    cin >> ch;
    if (ch == 'a')
      v.push_back(-1);
    else {
      if (v.empty() || v.back() == -1)
        v.push_back(k);
      else
        v.back() += k;
    }
  }
  a[v.size()] = 1;
  for (long long i = v.size() - 1; i >= 0; i--) {
    if (v[i] == -1) {
      a[i] = a[i + 1];
      continue;
    }
    if ((double)a[i + 1] * (v[i] + 1) > 2e18)
      a[i] = 2e18;
    else
      a[i] = a[i + 1] * (v[i] + 1);
  }
  for (long long i = 0; i < v.size(); i++) {
    if (v[i] == -1)
      cout << "a";
    else {
      while (x > a[i + 1]) cout << "b", x -= a[i + 1];
    }
  }
  cout << "\n";
}
signed main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}

### Prompt

Generate 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 MAX = 3e5 + 10;
const long long Time = 86400;
const long long X = 131;
const long long inf = 0x3f3f3f3f;
const double PI = 1e-4;
double pai = 3.14159265358979323846;
long long T, n, m, k, x, ans;
string s;
long long a[MAX];
vector<long long> v;
char ch;
void solve() {
  v.clear();
  cin >> n >> k >> x;
  for (long long i = 1; i <= n; i++) {
    cin >> ch;
    if (ch == 'a')
      v.push_back(-1);
    else {
      if (v.empty() || v.back() == -1)
        v.push_back(k);
      else
        v.back() += k;
    }
  }
  a[v.size()] = 1;
  for (long long i = v.size() - 1; i >= 0; i--) {
    if (v[i] == -1) {
      a[i] = a[i + 1];
      continue;
    }
    if ((double)a[i + 1] * (v[i] + 1) > 2e18)
      a[i] = 2e18;
    else
      a[i] = a[i + 1] * (v[i] + 1);
  }
  for (long long i = 0; i < v.size(); i++) {
    if (v[i] == -1)
      cout << "a";
    else {
      while (x > a[i + 1]) cout << "b", x -= a[i + 1];
    }
  }
  cout << "\n";
}
signed main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}

```