#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
using i32 = int;
const int INF = 0x3f3f3f3f;
const double tol = 1e-8;
const int P = 998244353;
int pop_count(i64 x) { return __builtin_popcountll(x); }
int pop_count(int x) { return __builtin_popcount(x); }
template <typename S, typename T>
bool check_and_save_min_in_lhs(S& lhs, const T& rhs) {
  return lhs < rhs ? 1 : lhs = rhs, 0;
}
template <typename S, typename T>
bool check_and_save_max_in_lhs(S& lhs, const T& rhs) {
  return lhs > rhs ? 1 : lhs = rhs, 0;
}
template <typename T>
T min_all(T ele) {
  return ele;
}
template <typename T, typename... Args>
T min_all(T fst_el, Args... others) {
  return min_all(fst_el, mins(others...));
}
template <typename T>
T maxs(T ele) {
  return ele;
}
template <typename T, typename... Args>
T maxs(T fst_el, Args... others) {
  return max(fst_el, maxs(others...));
}
template <typename T>
void read_int(T& num) {
  bool positive = true;
  char ch;
  num = 0;
  while (isspace(ch = getchar()))
    ;
  if (ch == '-')
    positive = false;
  else
    num = ch - '0';
  while ((ch = getchar()) >= '0' && ch <= '9') num = num * 10 + ch - '0';
  if (!positive) num = -num;
}
void read_ints(){};
template <typename T, typename... Args>
void read_ints(T& fst_el, Args&... others) {
  read_int(fst_el);
  read_ints(others...);
}
template <typename T>
void int_out(T num) {
  if (num < 0) {
    putchar('-');
    num = -num;
  }
  if (num > 9) int_out(num / 10);
  putchar(num % 10 + '0');
}
void int_outs() { puts(""); }
template <typename T, typename... Args>
void int_outs(T fst_ele, Args... others) {
  int_out(fst_ele);
  putchar(' ');
  int_outs(others...);
}
const int N = 2002;
char s[N];
int main() {
  int t;
  read_int(t);
  while (t--) {
    int n, k;
    i64 x;
    read_ints(n, k);
    read_int(x);
    gets(s);
    int len = 0, cur = 0;
    char ch[3] = "a*";
    vector<int> r[2];
    for (int i = 0, len; i < n; i += len) {
      len = 0;
      while (i + len < n && s[i + len] == ch[cur]) {
        ++len;
      }
      r[cur].push_back(len);
      cur ^= 1;
    }
    if (r[0].size() != r[1].size()) r[1].push_back(0);
    vector<i64> acc(r[1].size(), 0);
    int idx;
    i64 cnt = 1;
    for (idx = r[1].size() - 1; idx >= 0; --idx) {
      acc[idx] = cnt;
      if ((i64(k) * r[1][idx] + 1) > x / cnt) {
        break;
      } else {
        cnt = cnt * (1 + (i64)k * r[1][idx]);
      }
    }
    for (int i = 0; i < r[0].size(); ++i) {
      while (r[0][i]--) {
        putchar('a');
      }
      if (i >= idx && x > 0) {
        cnt = (x - 1) / acc[i];
        x -= cnt * acc[i];
        for (int j = 0; j < cnt; ++j) {
          putchar('b');
        }
      }
    }
    puts("");
  }
  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;
using i64 = long long;
using i32 = int;
const int INF = 0x3f3f3f3f;
const double tol = 1e-8;
const int P = 998244353;
int pop_count(i64 x) { return __builtin_popcountll(x); }
int pop_count(int x) { return __builtin_popcount(x); }
template <typename S, typename T>
bool check_and_save_min_in_lhs(S& lhs, const T& rhs) {
  return lhs < rhs ? 1 : lhs = rhs, 0;
}
template <typename S, typename T>
bool check_and_save_max_in_lhs(S& lhs, const T& rhs) {
  return lhs > rhs ? 1 : lhs = rhs, 0;
}
template <typename T>
T min_all(T ele) {
  return ele;
}
template <typename T, typename... Args>
T min_all(T fst_el, Args... others) {
  return min_all(fst_el, mins(others...));
}
template <typename T>
T maxs(T ele) {
  return ele;
}
template <typename T, typename... Args>
T maxs(T fst_el, Args... others) {
  return max(fst_el, maxs(others...));
}
template <typename T>
void read_int(T& num) {
  bool positive = true;
  char ch;
  num = 0;
  while (isspace(ch = getchar()))
    ;
  if (ch == '-')
    positive = false;
  else
    num = ch - '0';
  while ((ch = getchar()) >= '0' && ch <= '9') num = num * 10 + ch - '0';
  if (!positive) num = -num;
}
void read_ints(){};
template <typename T, typename... Args>
void read_ints(T& fst_el, Args&... others) {
  read_int(fst_el);
  read_ints(others...);
}
template <typename T>
void int_out(T num) {
  if (num < 0) {
    putchar('-');
    num = -num;
  }
  if (num > 9) int_out(num / 10);
  putchar(num % 10 + '0');
}
void int_outs() { puts(""); }
template <typename T, typename... Args>
void int_outs(T fst_ele, Args... others) {
  int_out(fst_ele);
  putchar(' ');
  int_outs(others...);
}
const int N = 2002;
char s[N];
int main() {
  int t;
  read_int(t);
  while (t--) {
    int n, k;
    i64 x;
    read_ints(n, k);
    read_int(x);
    gets(s);
    int len = 0, cur = 0;
    char ch[3] = "a*";
    vector<int> r[2];
    for (int i = 0, len; i < n; i += len) {
      len = 0;
      while (i + len < n && s[i + len] == ch[cur]) {
        ++len;
      }
      r[cur].push_back(len);
      cur ^= 1;
    }
    if (r[0].size() != r[1].size()) r[1].push_back(0);
    vector<i64> acc(r[1].size(), 0);
    int idx;
    i64 cnt = 1;
    for (idx = r[1].size() - 1; idx >= 0; --idx) {
      acc[idx] = cnt;
      if ((i64(k) * r[1][idx] + 1) > x / cnt) {
        break;
      } else {
        cnt = cnt * (1 + (i64)k * r[1][idx]);
      }
    }
    for (int i = 0; i < r[0].size(); ++i) {
      while (r[0][i]--) {
        putchar('a');
      }
      if (i >= idx && x > 0) {
        cnt = (x - 1) / acc[i];
        x -= cnt * acc[i];
        for (int j = 0; j < cnt; ++j) {
          putchar('b');
        }
      }
    }
    puts("");
  }
  return 0;
}

```