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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

int main() {
  long long n, m; cin >> n >> m;
  vector<long long> a(n);
  for(int i = 0; i < n; i++){
    cin >> a.at(i);
  }
  sort(a.rbegin(), a.rend());
  vector<vector<int>> mat(n, vector<int>(n));
  vector<long long> ind(n, -1);
  mat.at(0).at(0) = 0; ind.at(0) = 1;
  long long out = a.at(0);
  long long count = 1;
  while(count < m){
    long long tmp = a.at(0) + a.at(ind.at(0));
    long long ai = 0;
    int flag = 0;
    for(int i = 1; i < n; i++){
      if(ind.at(i) == -1) break;
      if(ind.at(i) == n) continue;
      if(a.at(i) + a.at(ind.at(i)) > tmp){
        tmp = a.at(i) + a.at(ind.at(i));
        ai = i;
        if(ind.at(i) == i) flag = 1;
      }
    }
    out += tmp;
    count++;
    ind.at(ind.at(ai))++;
    if(!flag){
      if(count < m){
        out += tmp;
        count++;
        ind.at(ai)++;
      }
    }
  }
  cout << out;
}