#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
int ioo = 1000000000;
long long loo = 1000000000000000000;
void add_self(int &x, int y) {
  if ((x += y) >= mod) x -= mod;
}
int add(int x, int y) { return add_self(x, y), x; }
void sub_self(int &x, int y) {
  if ((x -= y) < 0) x += mod;
}
int sub(int x, int y) { return sub_self(x, y), x; }
int mul(int x, int y) { return (long long)x * y % mod; }
long long po(long long k, long long n) {
  long long res = 1;
  while (n) {
    if (n % 2) res *= k;
    k *= k;
    n /= 2;
  }
  return res;
}
long long pom(long long k, long long n) {
  long long res = 1;
  while (n) {
    if (n % 2) res = mul(res, k);
    k = mul(k, k);
    n /= 2;
  }
  return res;
}
char a[2005];
vector<long long> v;
long long n, k, x;
int main() {
  int tt;
  scanf("%d", &tt);
  while (tt--) {
    v.clear();
    scanf("%lld%lld%lld", &n, &k, &x);
    scanf("%s", a);
    long long cnt = 0;
    for (int i = 0; i < n; i++) {
      if (a[i] == 'a') {
        if (cnt) v.push_back(cnt * k);
        cnt = 0;
      } else
        cnt++;
    }
    if (cnt) v.push_back(cnt * k);
    vector<long long> b = v;
    if (!b.empty()) b.back()++;
    int idx = b.size();
    for (int i = (int)b.size() - 2; i >= 1; i--) {
      if (b[i] + 1 <= (loo * 2) / b[i + 1])
        b[i] = (b[i] + 1) * b[i + 1];
      else {
        idx = i;
        break;
      }
    }
    if (idx != b.size())
      for (int i = idx; i >= 0; i--) b[i] = loo * 2;
    deque<long long> ans(v.size(), 0);
    for (int i = 0; i < v.size(); i++) {
      if (i == (int)v.size() - 1)
        ans[i] = x - 1;
      else {
        long long steps = min((x - 1) / b[i + 1], v[i]);
        ans[i] = steps;
        x -= steps * b[i + 1];
      }
    }
    for (int i = 0; i < n; i++) {
      if (a[i] == 'a')
        printf("a");
      else {
        if (i == 0 || a[i] != a[i - 1]) {
          long long cur = ans.front();
          while (cur--) printf("b");
          ans.pop_front();
        }
      }
    }
    printf("\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;
const int mod = 1000000007;
int ioo = 1000000000;
long long loo = 1000000000000000000;
void add_self(int &x, int y) {
  if ((x += y) >= mod) x -= mod;
}
int add(int x, int y) { return add_self(x, y), x; }
void sub_self(int &x, int y) {
  if ((x -= y) < 0) x += mod;
}
int sub(int x, int y) { return sub_self(x, y), x; }
int mul(int x, int y) { return (long long)x * y % mod; }
long long po(long long k, long long n) {
  long long res = 1;
  while (n) {
    if (n % 2) res *= k;
    k *= k;
    n /= 2;
  }
  return res;
}
long long pom(long long k, long long n) {
  long long res = 1;
  while (n) {
    if (n % 2) res = mul(res, k);
    k = mul(k, k);
    n /= 2;
  }
  return res;
}
char a[2005];
vector<long long> v;
long long n, k, x;
int main() {
  int tt;
  scanf("%d", &tt);
  while (tt--) {
    v.clear();
    scanf("%lld%lld%lld", &n, &k, &x);
    scanf("%s", a);
    long long cnt = 0;
    for (int i = 0; i < n; i++) {
      if (a[i] == 'a') {
        if (cnt) v.push_back(cnt * k);
        cnt = 0;
      } else
        cnt++;
    }
    if (cnt) v.push_back(cnt * k);
    vector<long long> b = v;
    if (!b.empty()) b.back()++;
    int idx = b.size();
    for (int i = (int)b.size() - 2; i >= 1; i--) {
      if (b[i] + 1 <= (loo * 2) / b[i + 1])
        b[i] = (b[i] + 1) * b[i + 1];
      else {
        idx = i;
        break;
      }
    }
    if (idx != b.size())
      for (int i = idx; i >= 0; i--) b[i] = loo * 2;
    deque<long long> ans(v.size(), 0);
    for (int i = 0; i < v.size(); i++) {
      if (i == (int)v.size() - 1)
        ans[i] = x - 1;
      else {
        long long steps = min((x - 1) / b[i + 1], v[i]);
        ans[i] = steps;
        x -= steps * b[i + 1];
      }
    }
    for (int i = 0; i < n; i++) {
      if (a[i] == 'a')
        printf("a");
      else {
        if (i == 0 || a[i] != a[i - 1]) {
          long long cur = ans.front();
          while (cur--) printf("b");
          ans.pop_front();
        }
      }
    }
    printf("\n");
  }
  return 0;
}

```