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<i32> = s.map(|a| a.parse().unwrap()).collect();
	a.sort();
	let mut l = 2;
	let mut r = 200001;
	let pattern = |n| a.iter().fold(0, |r, b| r + a.len() - a.lower_bound(&(n - *b)));
	while r - l > 1 {
		let mid = l + (r - l) / 2;
		if pattern(mid) >= m {
			l = mid;
		} else {
			r = mid;
		}
	}
	let pat = pattern(l);
	let mut cum = vec![0; n + 1];
	for i in 0..a.len() {
		cum[i + 1] = cum[i] + a[a.len() - 1 - i] as i64;
	}
	let ans = a.iter().fold(0, |r, b| {
		let c = a.len() - a.lower_bound(&(l - *b));
		r + c as i64 * *b as i64 + cum[c]
	}) - (pat - m) as i64 * l as i64;
	println!("{}", ans);
}

pub trait BinarySearch<'a, T: 'a> {
	fn lower_bound(&self, x: &T) -> usize;
}

impl<'a, T: 'a + Ord> BinarySearch<'a, T> for [T] {
	fn lower_bound(&self, x: &T) -> usize {
		let mut first = 0;
		let mut len = self.len() - first;
		while len > 0 {
			let half = len >> 1;
			let mid = first + half;
			use std::cmp::Ordering;
			match self[mid].cmp(x) {
				Ordering::Less => {
					first = mid + 1;
					len = len - half - 1;
				}
				Ordering::Equal | Ordering::Greater => {
					len = half;
				}
			}
		}
		first
	}
}