#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
long long INF = 1e18;
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long nCr(long long n, long long r) {
  long long res = 1;
  if (n < r) return -1;
  r = min(r, n - r);
  for (long long i = 0; i < r; i++) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
long long nPr(long long n, long long r) {
  long long res = 1;
  if (n < r) return -1;
  for (long long i = 0; i < r; i++) res *= (n - i);
  return res;
}
long long modexp(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    x = (x * x) % mod;
    y = y >> 1;
  }
  return res;
}
bool isOverflow(long long a, long long b) {
  if (a == 0 || b == 0) return false;
  long long result = a * b;
  if (a == result / b)
    return false;
  else
    return true;
}
void solve() {
  long long n, k, x;
  cin >> n >> k >> x;
  string s;
  cin >> s;
  vector<long long> v;
  for (long long i = 0; i < n; i++) {
    long long c = 0;
    while (i < n && s[i] == '*') i++, c++;
    if (c > 0) v.push_back(c), i--;
  }
  if (v.size() == 0) {
    cout << s << "\n";
    return;
  }
  long long curp = 0;
  for (long long i = 0; i < v.size(); i++) {
    v[i] = 1 + v[i] * k;
  }
  long long id = v.size() - 1;
  long long flag = 0;
  while (curp < x) {
    if ((double)v[id] > (double)INF / curp) {
      flag = 1;
      id--;
      break;
    }
    if (curp == 0)
      curp = v[id];
    else
      curp = curp * v[id];
    id--;
  }
  id++;
  if (flag == 0) curp = curp / v[id];
  vector<long long> b(v.size(), 0);
  long long y = x;
  for (long long i = id; i < v.size(); i++) {
    if (i == v.size() - 1) {
      b[i] = y;
      break;
    }
    b[i] = (y - 1) / (curp);
    y = y % curp;
    if (y == 0) y = curp;
    if (i + 1 < v.size()) curp = curp / v[i + 1];
  }
  b[b.size() - 1]--;
  string ans = "";
  long long ind = 0;
  for (long long i = 0; i < n; i++) {
    if (s[i] == 'a')
      ans.push_back('a');
    else {
      for (long long j = 0; j < b[ind]; j++) ans.push_back('b');
      ind++;
      while (i < n && s[i] == '*') i++;
      i--;
    }
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  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;
long long mod = 1e9 + 7;
long long INF = 1e18;
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long nCr(long long n, long long r) {
  long long res = 1;
  if (n < r) return -1;
  r = min(r, n - r);
  for (long long i = 0; i < r; i++) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
long long nPr(long long n, long long r) {
  long long res = 1;
  if (n < r) return -1;
  for (long long i = 0; i < r; i++) res *= (n - i);
  return res;
}
long long modexp(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    x = (x * x) % mod;
    y = y >> 1;
  }
  return res;
}
bool isOverflow(long long a, long long b) {
  if (a == 0 || b == 0) return false;
  long long result = a * b;
  if (a == result / b)
    return false;
  else
    return true;
}
void solve() {
  long long n, k, x;
  cin >> n >> k >> x;
  string s;
  cin >> s;
  vector<long long> v;
  for (long long i = 0; i < n; i++) {
    long long c = 0;
    while (i < n && s[i] == '*') i++, c++;
    if (c > 0) v.push_back(c), i--;
  }
  if (v.size() == 0) {
    cout << s << "\n";
    return;
  }
  long long curp = 0;
  for (long long i = 0; i < v.size(); i++) {
    v[i] = 1 + v[i] * k;
  }
  long long id = v.size() - 1;
  long long flag = 0;
  while (curp < x) {
    if ((double)v[id] > (double)INF / curp) {
      flag = 1;
      id--;
      break;
    }
    if (curp == 0)
      curp = v[id];
    else
      curp = curp * v[id];
    id--;
  }
  id++;
  if (flag == 0) curp = curp / v[id];
  vector<long long> b(v.size(), 0);
  long long y = x;
  for (long long i = id; i < v.size(); i++) {
    if (i == v.size() - 1) {
      b[i] = y;
      break;
    }
    b[i] = (y - 1) / (curp);
    y = y % curp;
    if (y == 0) y = curp;
    if (i + 1 < v.size()) curp = curp / v[i + 1];
  }
  b[b.size() - 1]--;
  string ans = "";
  long long ind = 0;
  for (long long i = 0; i < n; i++) {
    if (s[i] == 'a')
      ans.push_back('a');
    else {
      for (long long j = 0; j < b[ind]; j++) ans.push_back('b');
      ind++;
      while (i < n && s[i] == '*') i++;
      i--;
    }
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

```