#include <bits/stdc++.h>
using namespace std;
const int maxn = 11;
int t, n, k, a[maxn];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> t;
  while (t--) {
    cin >> n >> k;
    for (int i = 1; i <= n; i++) cin >> a[i];
    int s = 0, d = 0, x;
    for (x = 2; x <= n; x++) {
      d = pow(10, a[x] - a[x - 1]) - 1;
      s += d;
      if (k < s) {
        s -= d;
        cout << k - s + 1;
        for (int j = 1; j <= a[x - 1]; j++) cout << 9;
        cout << endl;
        break;
      }
    }
    if (x > n) {
      cout << k - s + 1;
      for (int i = 1; i <= a[n]; i++) cout << 9;
      cout << endl;
    }
  }
  return 0;
}

### Prompt

Develop a solution in CPP 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;
const int maxn = 11;
int t, n, k, a[maxn];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> t;
  while (t--) {
    cin >> n >> k;
    for (int i = 1; i <= n; i++) cin >> a[i];
    int s = 0, d = 0, x;
    for (x = 2; x <= n; x++) {
      d = pow(10, a[x] - a[x - 1]) - 1;
      s += d;
      if (k < s) {
        s -= d;
        cout << k - s + 1;
        for (int j = 1; j <= a[x - 1]; j++) cout << 9;
        cout << endl;
        break;
      }
    }
    if (x > n) {
      cout << k - s + 1;
      for (int i = 1; i <= a[n]; i++) cout << 9;
      cout << endl;
    }
  }
  return 0;
}

```