#include <bits/stdc++.h>
using namespace std;
long long a[15], b[15];
long long lt(long long x, long long y) {
  if (y == 0) return 1;
  long long ans = 1;
  for (long long i = 1; i <= y; i++) {
    ans *= x;
  }
  return ans;
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, k, max = 0, ans = 0;
    cin >> n >> k;
    b[n] = 0;
    for (long long i = 1; i <= n; i++) {
      cin >> a[i];
      if (i >= 2) {
        b[i - 1] = lt(10, a[i] - a[i - 1]) - 1;
        max += b[i - 1];
      }
    }
    if (max > k + 1) {
      for (long long i = n - 1; i >= 1; i--) {
        long long sub = min(max - k - 1, b[i]);
        max -= sub;
        b[i] -= sub;
        if (max == k + 1) {
          break;
        }
      }
    } else if (max <= k) {
      b[n] = k - max + 1;
    }
    for (long long i = 1; i <= n; i++) {
      ans += lt(10, a[i]) * b[i];
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Create a solution in cpp for the following problem:

In Berland, n different types of banknotes are used. Banknotes of the i-th type have denomination 10^{a_i} burles (burles are the currency used in Berland); the denomination of banknotes of the first type is exactly 1.

Let's denote f(s) as the minimum number of banknotes required to represent exactly s burles. For example, if the denominations of banknotes used in Berland are 1, 10 and 100, then f(59) = 14: 9 banknotes with denomination of 1 burle and 5 banknotes with denomination of 10 burles can be used to represent exactly 9 ⋅ 1 + 5 ⋅ 10 = 59 burles, and there's no way to do it with fewer banknotes.

For a given integer k, find the minimum positive number of burles s that cannot be represented with k or fewer banknotes (that is, f(s) > k).

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4) — number of test cases.

The first line of each test case contains two integers n and k (1 ≤ n ≤ 10; 1 ≤ k ≤ 10^9).

The next line contains n integers a_1, a_2, ..., a_n (0 = a_1 < a_2 < ... < a_n ≤ 9).

Output

For each test case, print one integer — the minimum positive number of burles s that cannot be represented with k or fewer banknotes.

Example

Input


4
3 13
0 1 2
2 777
0 4
3 255
0 1 3
10 1000000000
0 1 2 3 4 5 6 7 8 9


Output


59
778
148999
999999920999999999

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long a[15], b[15];
long long lt(long long x, long long y) {
  if (y == 0) return 1;
  long long ans = 1;
  for (long long i = 1; i <= y; i++) {
    ans *= x;
  }
  return ans;
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, k, max = 0, ans = 0;
    cin >> n >> k;
    b[n] = 0;
    for (long long i = 1; i <= n; i++) {
      cin >> a[i];
      if (i >= 2) {
        b[i - 1] = lt(10, a[i] - a[i - 1]) - 1;
        max += b[i - 1];
      }
    }
    if (max > k + 1) {
      for (long long i = n - 1; i >= 1; i--) {
        long long sub = min(max - k - 1, b[i]);
        max -= sub;
        b[i] -= sub;
        if (max == k + 1) {
          break;
        }
      }
    } else if (max <= k) {
      b[n] = k - max + 1;
    }
    for (long long i = 1; i <= n; i++) {
      ans += lt(10, a[i]) * b[i];
    }
    cout << ans << endl;
  }
  return 0;
}

```