#include <bits/stdc++.h>
using namespace std;
long long int power(long long int a, long long int b, long long int mod) {
  long long int ans = 1;
  while (b > 0) {
    if (b & 1) {
      ans = (ans * a) % mod;
    }
    a = (a * a) % mod;
    b = b / 2;
  }
  return ans;
}
int main() {
  vector<long long int> temp(11);
  temp[0] = 1;
  for (long long int i = 1; i <= 10; i++) temp[i] = temp[i - 1] * 10;
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    long long int n, k;
    cin >> n >> k;
    vector<long long int> arr(n);
    k++;
    for (long long int i = 0; i < n; i++) {
      cin >> arr[i];
    }
    long long int check = n - 1;
    long long int val = 0;
    for (long long int i = 0; i < n - 1; i++) {
      val += (temp[arr[i + 1] - arr[i]] - 1);
      if (val >= k) {
        check = i;
        break;
      }
    }
    long long int ans = 0;
    for (long long int j = 0; j <= check; j++) {
      if (j == check) {
        ans += (temp[arr[j]] * k);
        break;
      }
      long long int ta = (temp[arr[j + 1] - arr[j]] - 1);
      if (ta >= k) {
        ans += (temp[arr[j]] * k);
        break;
      } else {
        ans += (temp[arr[j]] * ta);
        k = k - ta;
      }
    }
    cout << ans << 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 int power(long long int a, long long int b, long long int mod) {
  long long int ans = 1;
  while (b > 0) {
    if (b & 1) {
      ans = (ans * a) % mod;
    }
    a = (a * a) % mod;
    b = b / 2;
  }
  return ans;
}
int main() {
  vector<long long int> temp(11);
  temp[0] = 1;
  for (long long int i = 1; i <= 10; i++) temp[i] = temp[i - 1] * 10;
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    long long int n, k;
    cin >> n >> k;
    vector<long long int> arr(n);
    k++;
    for (long long int i = 0; i < n; i++) {
      cin >> arr[i];
    }
    long long int check = n - 1;
    long long int val = 0;
    for (long long int i = 0; i < n - 1; i++) {
      val += (temp[arr[i + 1] - arr[i]] - 1);
      if (val >= k) {
        check = i;
        break;
      }
    }
    long long int ans = 0;
    for (long long int j = 0; j <= check; j++) {
      if (j == check) {
        ans += (temp[arr[j]] * k);
        break;
      }
      long long int ta = (temp[arr[j + 1] - arr[j]] - 1);
      if (ta >= k) {
        ans += (temp[arr[j]] * k);
        break;
      } else {
        ans += (temp[arr[j]] * ta);
        k = k - ta;
      }
    }
    cout << ans << endl;
  }
}

```