#include <bits/stdc++.h>
using namespace std;
void __print(long long x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  long long f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
vector<long long> factorise(long long x) {
  vector<long long> v;
  for (long long i = 1; i * i <= x; i++) {
    if (x % i == 0) {
      v.push_back(i);
      if (i * i != x) v.push_back(x / i);
    }
  }
  return v;
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x %= p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long nCk(long long x, long long k) {
  long long num = 1;
  for (long long i = 0; i < k; i++) {
    num *= (x - i) % 1000000007;
    num %= 1000000007;
  }
  long long den = 1;
  for (long long i = 1; i <= k; i++) {
    den *= i % 1000000007;
    den %= 1000000007;
  }
  return (num * power(den, 1000000007 - 2, 1000000007)) % 1000000007;
}
template <typename T>
bool is_palindrome(T v, long long n) {
  bool bl = true;
  for (long long i = 0; i < n / 2; i++) {
    if (v[i] != v[n - i - 1]) {
      bl = false;
      break;
    }
  }
  return bl;
}
bool cmp(vector<long long> &a, vector<long long> &b) {
  if (a[0] < b[0])
    return true;
  else if (a[0] > b[0])
    return false;
  else {
    if (a[1] - a[0] < b[1] - b[0]) return true;
    return false;
  }
  return false;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long test;
  cin >> test;
  long long te = 0;
  while (test--) {
    te++;
    long long n, k, x;
    cin >> n >> k >> x;
    vector<long long> v;
    v.push_back(1);
    string s;
    cin >> s;
    long long t = 0;
    for (long long i = n - 1; i >= 0; i--) {
      if (s[i] == '*') {
        t++;
      } else if (t) {
        if (v[v.size() - 1] > x / (t * k + 1))
          v.push_back(LLONG_MAX);
        else
          v.push_back(1ll * (t * k + 1) * v[v.size() - 1]);
        t = 0;
      }
    }
    if (t) {
      if (v[v.size() - 1] > x / (t * k + 1))
        v.push_back(LLONG_MAX);
      else
        v.push_back(1ll * (t * k + 1) * v[v.size() - 1]);
    };
    long long i = v.size() - 2;
    long long j = 0;
    x--;
    while (j < s.length()) {
      while (j < s.length() && s[j] != '*') {
        cout << s[j];
        j++;
      }
      if (j < s.length() && s[j] == '*') {
        long long temp = x / v[i];
        x = x % v[i];
        ;
        cout << string(temp, 'b');
        i--;
      }
      while (j < s.length() && s[j] == '*') j++;
    }
    cout << "\n";
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
void __print(long long x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  long long f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
vector<long long> factorise(long long x) {
  vector<long long> v;
  for (long long i = 1; i * i <= x; i++) {
    if (x % i == 0) {
      v.push_back(i);
      if (i * i != x) v.push_back(x / i);
    }
  }
  return v;
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x %= p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long nCk(long long x, long long k) {
  long long num = 1;
  for (long long i = 0; i < k; i++) {
    num *= (x - i) % 1000000007;
    num %= 1000000007;
  }
  long long den = 1;
  for (long long i = 1; i <= k; i++) {
    den *= i % 1000000007;
    den %= 1000000007;
  }
  return (num * power(den, 1000000007 - 2, 1000000007)) % 1000000007;
}
template <typename T>
bool is_palindrome(T v, long long n) {
  bool bl = true;
  for (long long i = 0; i < n / 2; i++) {
    if (v[i] != v[n - i - 1]) {
      bl = false;
      break;
    }
  }
  return bl;
}
bool cmp(vector<long long> &a, vector<long long> &b) {
  if (a[0] < b[0])
    return true;
  else if (a[0] > b[0])
    return false;
  else {
    if (a[1] - a[0] < b[1] - b[0]) return true;
    return false;
  }
  return false;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long test;
  cin >> test;
  long long te = 0;
  while (test--) {
    te++;
    long long n, k, x;
    cin >> n >> k >> x;
    vector<long long> v;
    v.push_back(1);
    string s;
    cin >> s;
    long long t = 0;
    for (long long i = n - 1; i >= 0; i--) {
      if (s[i] == '*') {
        t++;
      } else if (t) {
        if (v[v.size() - 1] > x / (t * k + 1))
          v.push_back(LLONG_MAX);
        else
          v.push_back(1ll * (t * k + 1) * v[v.size() - 1]);
        t = 0;
      }
    }
    if (t) {
      if (v[v.size() - 1] > x / (t * k + 1))
        v.push_back(LLONG_MAX);
      else
        v.push_back(1ll * (t * k + 1) * v[v.size() - 1]);
    };
    long long i = v.size() - 2;
    long long j = 0;
    x--;
    while (j < s.length()) {
      while (j < s.length() && s[j] != '*') {
        cout << s[j];
        j++;
      }
      if (j < s.length() && s[j] == '*') {
        long long temp = x / v[i];
        x = x % v[i];
        ;
        cout << string(temp, 'b');
        i--;
      }
      while (j < s.length() && s[j] == '*') j++;
    }
    cout << "\n";
  }
  return 0;
}

```