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: Accepted)

import java.util.*;
public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		long M = sc.nextLong();
		int A[] = new int[N];
		for(int i=0;i<N;i++)A[i]=sc.nextInt();
 		sc.close();
 		
 		rsort(A);
 		
 		//２数の和がKを越えるものだけ使うことにする。
 		int R = 2*100000 + 100;
 		int L = 0;
 		while(L+1<R) {		
 			int C = (L+R)/2;//このcが実現できるか調べるK
 			long subsum = 0;
 			for(int i=0;i<N;i++) {
 				int base = A[i];
 				int r = N;
 				int l = i-1;
 				while(l+1<r) {
 					int c = (l + r)/2;
 					if(base + A[c]>=C)l=c;
 					else r = c;
 				}
 				if(l-i<0)continue;
 				subsum += 1 + 2*(l-i);
 			}
 			if(subsum>=M) L = C;
 			else R = C;
 		}
 		//Lが最小のX
 		
 		long memosum = 0;
 		int min = Integer.MAX_VALUE;
 		int memo[] = new int[N];
 		for(int i=0;i<N;i++) {
				int base = A[i];
				int r = N;
				int l = i-1;
				while(l+1<r) {
					int c = (l + r)/2;
					if(base + A[c]>=L)l=c;
					else r = c;
				}
				memo[i]=l-i; //到着要素番号が入ってる
				
				if(l-i>=0) {
					memosum += 2*(l-i)+1;
					min = Math.min(min,base+A[l]);
				}
				
			}
 		
 		
 		
 		long sum[] = new long[N+1];
 		for(int i=0;i<N;i++)sum[i+1]=sum[i]+ A[i];
 		
 		long ans = 0;
 		for(int i=0;i<N;i++) {
 			if(memo[i]<0)break;
			int last = i+memo[i];
			long tmp = ((long)last-i+1)*2*A[i]+2*(sum[last+1]-sum[i+1]);
			ans += tmp;
 		}
 		long mul = Math.max(0,memosum-M);
 		ans -= mul*min;
 		System.out.println(ans);
	}
	
	static void rsort(int[] x) {
		Arrays.sort(x);
		int len = x.length;
		for(int i=0;i<len/2;i++) {
			int tmp = x[i];
			x[i]=x[len-1-i];
			x[len-1-i]=tmp;
		}
	}
}