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.Arrays;
import java.util.Scanner;

public class Main {
	static long can(long X , long A[] ){
		int N = A.length;
		int S = 0;
		for(int i = 0 ; i < N ; ++i){
			long Y = X - A[i];
			int idx = Arrays.binarySearch(A, Y);
			if(idx < 0){
				idx = - idx - 1;
			}
			S += N - idx;
		}
		return S;
	}
	static long solve(long X , long A[]){
		int N = A.length;
		long S[] = new long[N + 1];
		for(int i = 0 ; i < N ; ++i){
			S[i + 1] = S[i] + A[i];
		}
		long ret = 0;
		for(int i = 0 ; i < N ; ++i){
			long Y = X - A[i];
			int idx = Arrays.binarySearch(A, Y);
			if(idx < 0){
				idx = - idx - 1;
			}
			ret += S[N] - S[idx] + (N - idx) * A[i];
		}
		return ret;
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		long M = sc.nextLong();
		long A[] = new long[N];
		for(int i = 0 ; i < N ; ++i){
			A[i] = sc.nextLong();
		}
		Arrays.sort(A);
		long low = 0;
		long high = A[N - 1] * 2 + 1;
		while(low < high){
			long mid = (low + high) / 2;
			long c = can(mid , A);
			if(c <= M){
				high = mid;
			}else{
				low = mid + 1;
			}
		}
		long c = can(high, A);
		if(c == M){
			System.out.println(solve(high, A));			
		}else{
			long S = solve(high, A);
//			System.out.println(S+" "+c+" "+high);
			S += (high - 1) * (M - c);
			System.out.println(S);
		}
	}
}