#include <bits/stdc++.h>
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("O3")
#pragma GCC target("avx2")
using namespace std;
const long long N = 2e5 + 1, mod = 1e9 + 7;
void solve() {
  long long n, k;
  cin >> n >> k;
  k++;
  vector<long long> a(n);
  for (long long i = 0; i < n; i += 1) {
    cin >> a[i];
  }
  if (n == 1) {
    cout << k << "\n";
    return;
  }
  long long sum = 0, ans = 0;
  for (long long i = 1; i < n; i += 1) {
    long long p = pow(10, a[i] - a[i - 1]) - 1;
    if (sum + p < k) {
      sum += p;
      ans += p * pow(10, a[i - 1]);
    } else {
      ans += (k - sum) * pow(10, a[i - 1]);
      cout << ans << "\n";
      return;
    }
  }
  cout << (k - sum) << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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("unroll-loops")
#pragma GCC optimize("O3")
#pragma GCC target("avx2")
using namespace std;
const long long N = 2e5 + 1, mod = 1e9 + 7;
void solve() {
  long long n, k;
  cin >> n >> k;
  k++;
  vector<long long> a(n);
  for (long long i = 0; i < n; i += 1) {
    cin >> a[i];
  }
  if (n == 1) {
    cout << k << "\n";
    return;
  }
  long long sum = 0, ans = 0;
  for (long long i = 1; i < n; i += 1) {
    long long p = pow(10, a[i] - a[i - 1]) - 1;
    if (sum + p < k) {
      sum += p;
      ans += p * pow(10, a[i - 1]);
    } else {
      ans += (k - sum) * pow(10, a[i - 1]);
      cout << ans << "\n";
      return;
    }
  }
  cout << (k - sum) << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

```