#include <bits/stdc++.h>
using namespace std;
long long T, k, n, a[24], las, now, sum;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> T;
  while (T--) {
    cin >> k >> n;
    n++;
    memset(a, 0, sizeof(a));
    for (long long i = 1; i <= k; ++i) {
      long long f;
      cin >> f;
      a[f] = true;
    }
    if (n <= 9) {
      cout << n << endl;
      continue;
    }
    las = 0;
    now = 9;
    sum = 0;
    for (long long i = 1; i <= 19; ++i)
      if (a[i] == 0) {
        now = now * 10 + 9;
        if (now + sum >= n) {
          cout << n - sum;
          for (long long j = las - 1; j >= 0; j--) cout << '9';
          break;
        }
      } else {
        las = i;
        sum += now;
        now = 9;
        if (sum + now >= n) {
          cout << n - sum;
          for (long long j = i - 1; j >= 0; j--) cout << 9;
          break;
        }
      }
    cout << endl;
  }
}

### 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;
long long T, k, n, a[24], las, now, sum;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> T;
  while (T--) {
    cin >> k >> n;
    n++;
    memset(a, 0, sizeof(a));
    for (long long i = 1; i <= k; ++i) {
      long long f;
      cin >> f;
      a[f] = true;
    }
    if (n <= 9) {
      cout << n << endl;
      continue;
    }
    las = 0;
    now = 9;
    sum = 0;
    for (long long i = 1; i <= 19; ++i)
      if (a[i] == 0) {
        now = now * 10 + 9;
        if (now + sum >= n) {
          cout << n - sum;
          for (long long j = las - 1; j >= 0; j--) cout << '9';
          break;
        }
      } else {
        las = i;
        sum += now;
        now = 9;
        if (sum + now >= n) {
          cout << n - sum;
          for (long long j = i - 1; j >= 0; j--) cout << 9;
          break;
        }
      }
    cout << endl;
  }
}

```