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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

public class Main {

	public static void main(String[] args) throws IOException {
		// N, Mを取得
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		String n_m = in.readLine();

		// 半角スペースで区切られたn, mを取得
		int spaceIndex = n_m.indexOf(" ");
		int m = Integer.parseInt(n_m.substring(spaceIndex + 1));

		// 各パワーを取得
		String powers = in.readLine();
		String[] powerArray = powers.split(" ");

		// 取得したパワーを昇順にソート
		List<Integer> powerListAsc = new ArrayList<>();
		for (String power : powerArray) {
			powerListAsc.add(Integer.parseInt(power));
		}
		Collections.sort(powerListAsc);

		// 取得したパワーを降順にソート
		List<Integer> powerList = new ArrayList<>();
		for (int i = 0; i < powerListAsc.size(); i++) {
			powerList.add(powerListAsc.get(powerListAsc.size() - (i + 1)));
		}

		/*
		 * 幸福度を最大にするため、握手は下記の順序で行う
		 * (1)両手でパワーが一番大きい人とする
		 * (2)右手でパワーが一番大きい人、左手でパワーが二番目に大きい人とする
		 * (3)左手でパワーが一番大きい人、右手でパワーが二番目に大きい人とする
		 * (4)両手でパワーが二番目に大きい人とする
		 * (5)右手でパワーが二番目に大きい人、左手でパワーが三番目に大きい人とする
		 * (6)左手でパワーが二番目に大きい人、右手でパワーが三番目に大きい人とする
		 *
		 */
		int happiness = 0;

		// 3セットをカウントする
		int counter = 1;
		// 握手をm回行い幸福度を上げる
		for (int i = 0; i < m; i++) {

			// 両手で握手したことがない一番強い人とその次に強い人のパワーを取得
			int power1 = powerList.get(i);
			int power2 = powerList.get(i + 1);

			// 1セット目は両手でパワーが一番大きい人とする
			if (counter == 1) {

				happiness += power1 + power1;

				// 2セット目は右手でパワーが一番大きい人、左手でパワーが二番目に大きい人とする
			} else if (counter == 2) {

				happiness += power1 + power2;

				// 3セット目は左手でパワーが一番大きい人、右手でパワーが二番目に大きい人とする
			} else if (counter == 3) {

				happiness += power2 + power1;

			} else {
				counter = 1;
				break;
			}
		}

		// 最終的な幸福度の最大値を出力
		System.out.println(happiness);
	}

}