#include <bits/stdc++.h>
using namespace std;
long long int power(long long int a, long long int b) {
  long long int result;
  if (b == 0) return 1;
  if (b % 2 != 0) {
    result = a;
    result = (result * (power(a, b - 1)));
  } else {
    result = power(a, b / 2);
    result = (result * result);
  }
  return (long long int)result;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, k;
    cin >> n >> k;
    long long int sum = 0;
    long long int prev = 0;
    vector<long long int> v(n);
    for (auto &i : v) cin >> i;
    bool f = 1;
    for (int i = 0; i < n - 1; i++) {
      long long int temp = power(10, v[i + 1] - v[i]) - 1;
      if (temp <= k) {
        sum += temp * (power(10, v[i]));
        k -= temp;
      } else {
        sum += (k + 1) * (power(10, v[i]));
        f = 0;
        break;
      }
    }
    if (f) {
      sum += (k + 1) * (power(10, v[n - 1]));
    }
    cout << sum << "\n";
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

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 result;
  if (b == 0) return 1;
  if (b % 2 != 0) {
    result = a;
    result = (result * (power(a, b - 1)));
  } else {
    result = power(a, b / 2);
    result = (result * result);
  }
  return (long long int)result;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, k;
    cin >> n >> k;
    long long int sum = 0;
    long long int prev = 0;
    vector<long long int> v(n);
    for (auto &i : v) cin >> i;
    bool f = 1;
    for (int i = 0; i < n - 1; i++) {
      long long int temp = power(10, v[i + 1] - v[i]) - 1;
      if (temp <= k) {
        sum += temp * (power(10, v[i]));
        k -= temp;
      } else {
        sum += (k + 1) * (power(10, v[i]));
        f = 0;
        break;
      }
    }
    if (f) {
      sum += (k + 1) * (power(10, v[n - 1]));
    }
    cout << sum << "\n";
  }
  return 0;
}

```