#include <bits/stdc++.h>
using namespace std;
int main(int argc, char const *argv[]) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    k++;
    vector<long long> arr(n, 0);
    for (long long i = 0; i < n; i++) {
      cin >> arr[i];
    }
    long long ans = 0;
    long long mine;
    for (long long i = 0; i < n - 1; i++) {
      long long temp = arr[i + 1] - arr[i];
      long long count = ceil(pow(10.0, temp)) - 1;
      if (k < count) {
        mine = k * ceil(pow(10.0, arr[i]));
        ans += mine;
        k = 0;
        break;
      } else {
        mine = count * ceil(pow(10.0, arr[i]));
        ans += mine;
        k -= count;
      }
    }
    mine = k * ceil(pow(10.0, arr[n - 1]));
    ans += mine;
    cout << ans << "\n";
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
int main(int argc, char const *argv[]) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    k++;
    vector<long long> arr(n, 0);
    for (long long i = 0; i < n; i++) {
      cin >> arr[i];
    }
    long long ans = 0;
    long long mine;
    for (long long i = 0; i < n - 1; i++) {
      long long temp = arr[i + 1] - arr[i];
      long long count = ceil(pow(10.0, temp)) - 1;
      if (k < count) {
        mine = k * ceil(pow(10.0, arr[i]));
        ans += mine;
        k = 0;
        break;
      } else {
        mine = count * ceil(pow(10.0, arr[i]));
        ans += mine;
        k -= count;
      }
    }
    mine = k * ceil(pow(10.0, arr[n - 1]));
    ans += mine;
    cout << ans << "\n";
  }
  return 0;
}

```