#include <bits/stdc++.h>
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long p10[10] = {};
  p10[0] = 1;
  for (int i = 1; i <= 9; ++i) p10[i] = p10[i - 1] * 10;
  int t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    k++;
    long long a[n], ans = 0;
    for (int i = 0; i <= n - 1; ++i) cin >> a[i];
    for (int i = 1; i <= n - 1; ++i) {
      long long take = min(k, p10[a[i] - a[i - 1]] - 1);
      ans += p10[a[i - 1]] * take;
      k -= take;
    }
    ans += p10[a[n - 1]] * k;
    cout << ans << "\n";
  }
  return 0;
}

### Prompt

Please formulate 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>
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long p10[10] = {};
  p10[0] = 1;
  for (int i = 1; i <= 9; ++i) p10[i] = p10[i - 1] * 10;
  int t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    k++;
    long long a[n], ans = 0;
    for (int i = 0; i <= n - 1; ++i) cin >> a[i];
    for (int i = 1; i <= n - 1; ++i) {
      long long take = min(k, p10[a[i] - a[i - 1]] - 1);
      ans += p10[a[i - 1]] * take;
      k -= take;
    }
    ans += p10[a[n - 1]] * k;
    cout << ans << "\n";
  }
  return 0;
}

```