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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Arrays;

public class Main {
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] sa = br.readLine().split(" ");
		int n = Integer.parseInt(sa[0]);
		long m = Long.parseLong(sa[1]);
		sa = br.readLine().split(" ");
		long[] a = new long[n];
		for (int i = 0; i < n; i++) {
			a[i] = Integer.parseInt(sa[i]);
		}
		br.close();

		Arrays.parallelSort(a);
		long ng = 1;
		long ok = a[n - 1] * 2 + 1;
		while (ok - ng > 1) {
			long mid = (ok + ng) / 2;
			long cnt = 0;
			for (int i = 0; i < n; i++) {
				int idx = binarySearch(a, mid - a[i] - 0.5);
				cnt += n - idx;
			}
			if (cnt < m) {
				ok = mid;
			} else {
				ng = mid;
			}
		}

		long[] b = new long[n + 1];
		for (int i = 0; i < n; i++) {
			b[i + 1] = b[i] + a[i];
		}

		long ans = 0;
		long cnt = 0;
		for (int i = 0; i < n; i++) {
			int idx = binarySearch(a, ng - a[i] - 0.5);
			long c = n - idx;
			cnt += c;
			long val = a[i] * c + (b[n] - b[idx]);
			ans += val;
		}
		ans -= (cnt - m) * ng;
		System.out.println(ans);
	}

	static int binarySearch(long[] array, double val) {
		int start = -1;
		int end = array.length;

		while (Math.abs(end - start) > 1) {
			int mid = (end + start) / 2;
			if (mid == array.length) {
				return mid;
			}
			if (array[mid] >= val) {
				end = mid;
			} else {
				start = mid;
			}
		}
		return end;
	}
}