#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    int num[100];
    for (int i = 0; i < n; i++) cin >> num[i];
    long long number[1000];
    memset(number, 0, sizeof(number));
    for (int i = 0; i < n; i++) number[i] = pow(10, num[i]);
    long long prime[1000];
    memset(prime, 0, sizeof(prime));
    for (int i = 1; i < n; i++)
      prime[i - 1] = (number[i] - number[i - 1]) / number[i - 1];
    long long sum = 0;
    int i = 0;
    while (i < n - 1 && sum <= k) {
      sum += prime[i++];
    }
    if (i == n - 1 && k >= sum) {
      int x = k - sum;
      long long sum1 = 0;
      for (int j = 0; j < n - 1; j++) {
        sum1 += number[j] * prime[j];
      }
      long long ans = sum1 + (x + 1) * number[n - 1];
      cout << ans << endl;
    } else {
      sum -= prime[--i];
      int x = k - sum;
      long long sum1 = 0;
      for (int j = 0; j < i; j++) {
        sum1 += number[j] * prime[j];
      }
      long long ans = sum1 + (x + 1) * number[i];
      cout << ans << endl;
    }
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    int num[100];
    for (int i = 0; i < n; i++) cin >> num[i];
    long long number[1000];
    memset(number, 0, sizeof(number));
    for (int i = 0; i < n; i++) number[i] = pow(10, num[i]);
    long long prime[1000];
    memset(prime, 0, sizeof(prime));
    for (int i = 1; i < n; i++)
      prime[i - 1] = (number[i] - number[i - 1]) / number[i - 1];
    long long sum = 0;
    int i = 0;
    while (i < n - 1 && sum <= k) {
      sum += prime[i++];
    }
    if (i == n - 1 && k >= sum) {
      int x = k - sum;
      long long sum1 = 0;
      for (int j = 0; j < n - 1; j++) {
        sum1 += number[j] * prime[j];
      }
      long long ans = sum1 + (x + 1) * number[n - 1];
      cout << ans << endl;
    } else {
      sum -= prime[--i];
      int x = k - sum;
      long long sum1 = 0;
      for (int j = 0; j < i; j++) {
        sum1 += number[j] * prime[j];
      }
      long long ans = sum1 + (x + 1) * number[i];
      cout << ans << endl;
    }
  }
  return 0;
}

```