#include <bits/stdc++.h>
using namespace std;
const long long M = 1e9 + 7;
const long long IM = 1e18 + 37;
const long long N = 2e5 + 10;
long long solve(long long cn, long long a[], long long n, long long k) {
  long long ans = 0;
  for (long long i = n - 1; i >= 0; i--) {
    long long val = 1, now = a[i];
    while (now--) val = val * 10;
    ans += cn / val;
    cn = cn - ((cn / val) * val);
  }
  return (ans > k);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    long long a[n];
    for (long long &it : a) cin >> it;
    long long ans = 0;
    long long val = 1;
    long long u = k;
    for (long long i = 0; i < n; i++) {
      long long id = 99999999999;
      if (i + 1 <= n - 1) {
        id = 0;
        long long cn = a[i + 1] - a[i];
        while (cn--) {
          id = id * 10 + 9;
        }
      }
      val = 1;
      long long now = a[i];
      while (now--) {
        val = val * 10;
      }
      if (solve(ans + (k + 1) * val, a, n, u)) {
        ans += (k + 1) * val;
        break;
      }
      id = min(k, id);
      k -= id;
      ans += id * val;
    }
    cout << ans << "\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;
const long long M = 1e9 + 7;
const long long IM = 1e18 + 37;
const long long N = 2e5 + 10;
long long solve(long long cn, long long a[], long long n, long long k) {
  long long ans = 0;
  for (long long i = n - 1; i >= 0; i--) {
    long long val = 1, now = a[i];
    while (now--) val = val * 10;
    ans += cn / val;
    cn = cn - ((cn / val) * val);
  }
  return (ans > k);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    long long a[n];
    for (long long &it : a) cin >> it;
    long long ans = 0;
    long long val = 1;
    long long u = k;
    for (long long i = 0; i < n; i++) {
      long long id = 99999999999;
      if (i + 1 <= n - 1) {
        id = 0;
        long long cn = a[i + 1] - a[i];
        while (cn--) {
          id = id * 10 + 9;
        }
      }
      val = 1;
      long long now = a[i];
      while (now--) {
        val = val * 10;
      }
      if (solve(ans + (k + 1) * val, a, n, u)) {
        ans += (k + 1) * val;
        break;
      }
      id = min(k, id);
      k -= id;
      ans += id * val;
    }
    cout << ans << "\n";
  }
  return 0;
}

```