Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

5 3
10 14 19 34 33

Sample Output 1

202

Let us say that Takahashi performs the following handshakes:

In the first handshake, Takahashi shakes the left hand of Guest 4 and the right hand of Guest 4.

In the second handshake, Takahashi shakes the left hand of Guest 4 and the right hand of Guest 5.

In the third handshake, Takahashi shakes the left hand of Guest 5 and the right hand of Guest 4.

Then, we will have the happiness of (34+34)+(34+33)+(33+34)=202.

We cannot achieve the happiness of 203 or greater, so the answer is 202.

Sample Input 2

9 14
1 3 5 110 24 21 34 5 3

Sample Output 2

1837

Sample Input 3

9 73
67597 52981 5828 66249 75177 64141 40773 79105 16076

Sample Output 3

8128170

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < n; i++)
using namespace std;
typedef long long int ll;

int main() {
  ll n, m; cin >> n >> m;
  vector<int> a(n); rep(i, n) cin >> a[i];
  sort(a.rbegin(), a.rend());
  ll ans = 0;
  for (ll i = 0; i < n; i++) {
    for (ll j = i; j < n; j++) {
      if (m > 0) {
        if (m-2 >= 0 && i != j) {
          // cout << m << " " << i << " " << j << " " << 2 * (a[i] + a[j]) << endl;
          ans += 2 * (a[i] + a[j]);
          m -= 2;
        }
        else {
          // cout << m  << " " << i << " " << j << " " << a[i] + a[j] << endl;
          ans += a[i] + a[j];
          m--;
        }
      }
      else {
        cout << ans << endl;
        return 0;
      }
    }
  }
  return 0;
}