#include <bits/stdc++.h>
using namespace std;
long long a[15], n, k, t;
vector<long long> ans;
long long ten[] = {1,       10,       100,       1000,       10000,      100000,
                   1000000, 10000000, 100000000, 1000000000, 10000000000};
signed main() {
  cin >> t;
  while (t--) {
    cin >> n >> k;
    k++;
    for (long long i = 1; i <= n; i++) cin >> a[i];
    ans.clear();
    for (long long i = 1; i <= n; i++)
      if (i < n) {
        long long d = a[i + 1] - a[i];
        if (k <= ten[d] - 1) {
          ans.push_back(k);
          break;
        } else {
          ans.push_back(ten[d] - 1);
          k -= ten[d] - 1;
        }
      } else if (i == n)
        ans.push_back(k);
    for (long long i = ans.size() - 1; i >= 0; i--) printf("%lld", ans[i]);
    putchar('\n');
  }
}

### 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;
long long a[15], n, k, t;
vector<long long> ans;
long long ten[] = {1,       10,       100,       1000,       10000,      100000,
                   1000000, 10000000, 100000000, 1000000000, 10000000000};
signed main() {
  cin >> t;
  while (t--) {
    cin >> n >> k;
    k++;
    for (long long i = 1; i <= n; i++) cin >> a[i];
    ans.clear();
    for (long long i = 1; i <= n; i++)
      if (i < n) {
        long long d = a[i + 1] - a[i];
        if (k <= ten[d] - 1) {
          ans.push_back(k);
          break;
        } else {
          ans.push_back(ten[d] - 1);
          k -= ten[d] - 1;
        }
      } else if (i == n)
        ans.push_back(k);
    for (long long i = ans.size() - 1; i >= 0; i--) printf("%lld", ans[i]);
    putchar('\n');
  }
}

```