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: Runtime Error)

import java.util.Arrays;
import java.util.Scanner;

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();
		Arrays.parallelSort(a);
		int max = a[n-1];

		//cumulative sum
		int[] csum = new int[max*2];
		int p = 0;
		for (int i = 0; i < csum.length; i++) {
			while(p < n && a[p] < i) {
				p++;
				if(p == n) break;
			}
			csum[i] = n-p;
		}

		//binary search
		int l = -1; int r = max*2+1;
		while(r-l> 1) {
			int mid = (r+l)/2;
			int sum = 0;
			for (int i = 0; i < n; i++) {
				int res = Math.max(mid-a[i], 0);
				sum += csum[res];
			}
			//System.out.printf("mid:%d, sum:%d%n", mid, sum);
			if(sum >= m) l = mid;
			else r = mid;
		}

		//cumulative sum
		int[] csum2 = new int[n];
		csum2[0] = a[0];
		for (int i = 1; i < n; i++) {
			csum2[i] = csum2[i-1] + a[i];
		}
		//System.out.println(Arrays.toString(csum2));
		int count = 0; int sum = 0;
		for (int i = 0; i < n; i++) {
			int res = Math.max(l-a[i]+1, 0);
			if(n == csum[res]) {
				sum += a[i]*csum[res] + csum2[n-1];
			}else {
				sum += a[i]*csum[res] + (csum2[n-1]-csum2[n-csum[res]-1]);
			}

			count += csum[res];
			//System.out.println(sum);
		}
		sum += (m-count)*l;
		System.out.println(sum);
		sc.close();
	}
}