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: Time Limit Exceeded)

#include<set>
#include<tuple>
#include<algorithm> // sort,reverse,min,lower_bound
#include<vector>
#include<iostream>
using namespace std;
#define df 0
#define MOD 1'000'000'007
#define rep(i,n) for(int i=0;i<(n);i++)
#define repa(i,a,n) for(int i=(a);i<(n);i++)
typedef long int li;


int main(){
  if(df) printf("*debug mode*\n");
  int n,m; cin >>n >>m;
  vector<int> a(n);
  for(int& x: a) cin >>x;
  sort(a.begin(),a.end());
  reverse(a.begin(),a.end());
  set<tuple<int,int,int>> mp;
  rep(i,n){
    rep(j,n){
      mp.insert(make_tuple(-(a.at(i)+a.at(j)),i,j));
    }
  }
   auto itr=mp.begin();
  li ans=0;
  rep(i,m){
       ans-=get<0>(*itr);
       itr++;
  }
  
  
  
  cout << ans;
}

/// confirm df==0 ///