#include <iostream>
#include <algorithm>
using namespace std;

int main() {
  long long n, m;
  cin >> n >> m;
  long long a[n + 1], b[n + 1];
  for (int i = 1; cin >> a[i]; i++) {
  }
  sort(a + 1, a + n + 1);
  b[0] = 0;
  for (long long i = 1; i <= n; i++) {
    b[i] = b[i - 1] + a[i];
  }
  long long left = a[1] * 2, right = a[n] * 2 + 1;
  auto count = [&](long long x) {
    pair<long long, long long> result = {0, 0};
    for (long long i = 1, j = n; j > 0; j--) {
      while (i <= n && a[i] + a[j] < x) {
        i++;
      }
      result.first += n - i + 1;
      result.second += b[n] - b[i - 1] + a[j] * (n - i + 1);
    }
    return result;
  };
  while (left < right) {
    long long mid = (left + right) / 2;
    if (count(mid).first <= m) {
      right = mid;
    } else {
      left = mid + 1;
    }
  }
  auto result = count(left);
  cout << result.second + (m - result.first) * (left - 1);
  return 0;
}
### Prompt

Create a solution in Cpp for the following problem:

Takahashi has come to a party as a special guest. There are N ordinary guests at the party. The i-th ordinary guest has a power of A_i.

Takahashi has decided to perform M handshakes to increase the happiness of the party (let the current happiness be 0). A handshake will be performed as follows:

* Takahashi chooses one (ordinary) guest x for his left hand and another guest y for his right hand (x and y can be the same).
* Then, he shakes the left hand of Guest x and the right hand of Guest y simultaneously to increase the happiness by A_x+A_y.



However, Takahashi should not perform the same handshake more than once. Formally, the following condition must hold:

* Assume that, in the k-th handshake, Takahashi shakes the left hand of Guest x_k and the right hand of Guest y_k. Then, there is no pair p, q (1 \leq p < q \leq M) such that (x_p,y_p)=(x_q,y_q).



What is the maximum possible happiness after M handshakes?

Constraints

* 1 \leq N \leq 10^5
* 1 \leq M \leq N^2
* 1 \leq A_i \leq 10^5
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N M
A_1 A_2 ... A_N


Output

Print the maximum possible happiness after M handshakes.

Examples

Input

5 3
10 14 19 34 33


Output

202


Input

9 14
1 3 5 110 24 21 34 5 3


Output

1837


Input

9 73
67597 52981 5828 66249 75177 64141 40773 79105 16076


Output

8128170

### Response

```cpp
#include <iostream>
#include <algorithm>
using namespace std;

int main() {
  long long n, m;
  cin >> n >> m;
  long long a[n + 1], b[n + 1];
  for (int i = 1; cin >> a[i]; i++) {
  }
  sort(a + 1, a + n + 1);
  b[0] = 0;
  for (long long i = 1; i <= n; i++) {
    b[i] = b[i - 1] + a[i];
  }
  long long left = a[1] * 2, right = a[n] * 2 + 1;
  auto count = [&](long long x) {
    pair<long long, long long> result = {0, 0};
    for (long long i = 1, j = n; j > 0; j--) {
      while (i <= n && a[i] + a[j] < x) {
        i++;
      }
      result.first += n - i + 1;
      result.second += b[n] - b[i - 1] + a[j] * (n - i + 1);
    }
    return result;
  };
  while (left < right) {
    long long mid = (left + right) / 2;
    if (count(mid).first <= m) {
      right = mid;
    } else {
      left = mid + 1;
    }
  }
  auto result = count(left);
  cout << result.second + (m - result.first) * (left - 1);
  return 0;
}
```