#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n, k;
  cin >> n >> k;
  long long ar[n];
  for (int i = 0; i < n; i++) {
    cin >> ar[i];
    ar[i] = pow(10, ar[i]);
  }
  sort(ar, ar + n);
  if (n == 1) {
    cout << k + 1 << endl;
    return;
  }
  if (k < ar[1] - 1) {
    cout << k + 1 << endl;
    return;
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    if (i + 1 < n) {
      long long kk = ar[i + 1] / ar[i];
      kk--;
      long long use = min(kk, k);
      if (i == 0) use--;
      k -= use;
      ans += ar[i] * use;
      if (k == 0) break;
    } else {
      ans += ar[i] * k;
      break;
    }
  }
  cout << ans + 1 << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

### 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;
void solve() {
  long long n, k;
  cin >> n >> k;
  long long ar[n];
  for (int i = 0; i < n; i++) {
    cin >> ar[i];
    ar[i] = pow(10, ar[i]);
  }
  sort(ar, ar + n);
  if (n == 1) {
    cout << k + 1 << endl;
    return;
  }
  if (k < ar[1] - 1) {
    cout << k + 1 << endl;
    return;
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    if (i + 1 < n) {
      long long kk = ar[i + 1] / ar[i];
      kk--;
      long long use = min(kk, k);
      if (i == 0) use--;
      k -= use;
      ans += ar[i] * use;
      if (k == 0) break;
    } else {
      ans += ar[i] * k;
      break;
    }
  }
  cout << ans + 1 << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

```