#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long int n, k;
  cin >> n >> k;
  vector<long long int> v(n);
  for (int i = 0; i < n; i++) cin >> v[i];
  ;
  sort(v.begin(), v.end());
  vector<long long int> p(10, 0);
  p[0] = 1;
  for (int i = 1; i <= 9; i++) {
    p[i] = p[i - 1] * 10;
  }
  long long int s = 2e18;
  long long int ans = 0;
  long long int cnt = 0;
  k++;
  for (int i = 0; i < n; i++) {
    if (i == n - 1 || ans + k * p[v[i]] < p[v[i + 1]]) {
      long long int val = ans + k * p[v[i]];
      s = min(val, s);
    }
    long long int diff = 0;
    if (i != n - 1) {
      diff = v[i + 1] - v[i];
    }
    ans += min(p[diff] - 1, k) * p[v[i]];
    k -= min(p[diff] - 1, k);
  }
  cout << s << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 int n, k;
  cin >> n >> k;
  vector<long long int> v(n);
  for (int i = 0; i < n; i++) cin >> v[i];
  ;
  sort(v.begin(), v.end());
  vector<long long int> p(10, 0);
  p[0] = 1;
  for (int i = 1; i <= 9; i++) {
    p[i] = p[i - 1] * 10;
  }
  long long int s = 2e18;
  long long int ans = 0;
  long long int cnt = 0;
  k++;
  for (int i = 0; i < n; i++) {
    if (i == n - 1 || ans + k * p[v[i]] < p[v[i + 1]]) {
      long long int val = ans + k * p[v[i]];
      s = min(val, s);
    }
    long long int diff = 0;
    if (i != n - 1) {
      diff = v[i + 1] - v[i];
    }
    ans += min(p[diff] - 1, k) * p[v[i]];
    k -= min(p[diff] - 1, k);
  }
  cout << s << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

```