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 C (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>
#include <stdlib.h>

int cmpint(const void* n1, const void* n2){
	if (*(int*)n1 > * (int*)n2) {
		return -1;//降順なら-1
	}
	else if (*(int*)n1 < *(int*)n2) {
		return 1;//降順なら1
	}
	else {
		return 0;
	}
}

int main() {
	int n;
	long long m;
	int ans=0;
	scanf("%d%ld", &n, &m);
	int a[100000]={};
	for (int i = 0; i < n; i++) {
		scanf("%d", &a[i]);
	}

	qsort(a, n, sizeof(int), cmpint);
	long long cou=0;
	int i=0;
	while(1) {
		ans+=a[i]+a[i];
		cou++;
		if(cou==m)
			break;
		ans+=a[i]+a[i+1];
		cou++;
		if(cou==m)
			break;
		ans+=a[i+1]+a[i];
		cou++;
		if(cou==m)
			break;
		i++;
	}
	printf("%d\n", ans);
	return 0;
}