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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
import java.math.*;
public class Main{
	public static void main(String[] args){
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		long m=sc.nextLong();
		double[] a=new double[n];
		long[] sum=new long[n+1];
		for(int i=0; i<n; i++){
			a[i]=sc.nextDouble();
		}
		Arrays.sort(a);
		sum[n-1]=(long)a[n-1];
		for(int i=n-2; i>=0; i--){
			sum[i]=sum[i+1]+(long)a[i];
		}
		long left=0;
		long right=200001;
		while(right-left>1){
			long mid=(left+right)/2;
			int count=0;
			for(int i=0; i<n; i++){
				int index=-Arrays.binarySearch(a,mid-a[i]-0.5)-1;
				count+=n-index;
			}
			if(count>=m){
				left=mid;
			}else{
				right=mid;
			}
		}
		long ans=0;
		for(int i=0; i<n; i++){
			int index=-Arrays.binarySearch(a,left-a[i]+0.5)-1;
			m-=n-index;
			ans+=a[i]*(n-index)+sum[index];
		}
		System.out.println(ans+m*left);
	}
}