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 <iostream>
#include <cmath>
#include <vector>
#include <map>
#include <stack>
#include <queue>
#include <set>
#include <algorithm>
#include <iomanip>
#include <string.h>

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define ALL(a)  (a).begin(),(a).end()

typedef long long lint;

using namespace std;

int main(){
  lint N,M;
  cin>>N>>M;
  vector<lint> A(N);
  REP(i,N)cin>>A[i];
  sort(ALL(A),greater<lint>());
  // for(auto e:A)cerr<<e<<" ";
  // cerr<<endl;
  lint S[N];
  S[0]=A[0];
  FOR(i,1,N)S[i]=S[i-1]+A[i];
  lint ok=2e5+1;
  lint ng=0;
  lint ans=0;
  while(ok-ng>1){
    lint c=(ok+ng)/2;
    lint score=0;
    lint num=0;
    // cerr<<c<<endl;
    REP(i,N){
      lint l=-1,r=N;
      while(r-l>1){
        int cc=(l+r)/2;
        if(A[cc]>=c-A[i])l=cc;
        else r=cc;
      }
      // cerr<<l<<endl;
      if(l>=0){
        score+=S[l]+(l+1)*A[i];
        num+=l+1;
      }
    }
    // cerr<<endl;
    if(num<=M){
      ans=score+(c-(lint)1)*(M-num);
      ok=c;
      // cerr<<c<<" "<<num<<endl;
    }else{
      ng=c;
    }
  }
  cout<<ans<<endl;
  return 0;
}