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 Rust (1.15.1) (judge verdict: Accepted)

fn main() {
	let mut s = String::new();
	use std::io::Read;
	std::io::stdin().read_to_string(&mut s).unwrap();
	let mut s = s.split_whitespace();
	let n: usize = s.next().unwrap().parse().unwrap();
	let m: usize = s.next().unwrap().parse().unwrap();
	let mut a: Vec<i64> = s.map(|a| a.parse().unwrap()).collect();
	a.sort();
	let mut cum = vec![0; n + 1];
	let mut pnum = vec![0; *a.last().unwrap() as usize * 2 + 1];
	for (i, a) in a.iter().rev().enumerate() {
		cum[i + 1] = cum[i] + *a;
		pnum[*a as usize] += 1;
	}
	for i in (0..pnum.len() - 1).rev() {
		pnum[i] += pnum[i + 1];
	}
	let mut l = 2;
	let mut r = a.last().unwrap() * 2 + 1;
	'a: while r - l > 1 {
		let mid = l + (r - l) / 2;
		let mut cnt = 0;
		for a in a.iter().rev() {
			cnt += pnum[std::cmp::max(0, mid - *a) as usize];
			if cnt >= m {
				l = mid;
				continue 'a;
			}
		}
		r = mid;
	}
	let (lsum, lcnt) = a.iter().fold((0, 0), |r, a| {
		let c = pnum[std::cmp::max(0, l - *a) as usize];
		(r.0 + c as i64 * *a + cum[c], r.1 + c)
	});
	let ans = lsum - (lcnt - m) as i64 * l;
	println!("{}", ans);
}