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)

// This code is generated by [cargo-atcoder](https://github.com/tanakh/cargo-atcoder)
// Original source code:
/*
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;

fn main() {
    input! {
        n: usize,
        mut m : usize,
        mut a: [i64; n],
    }
    a.sort_by_key(|x| std::cmp::Reverse(*x));
    let mut s = vec![0; n + 1];
    for i in 0..n {
        s[i + 1] = s[i] + a[i];
    }

    let x = kyopro_lib::misc::lower_bound(0, 200_000, |x| {
        (0..n)
            .map(|i| kyopro_lib::misc::lower_bound(0, n, |j| a[i] + a[j] <= x))
            .sum::<usize>()
            <= m
    });

    let res = (0..n)
        .map(|i| (i, kyopro_lib::misc::lower_bound(0, n, |j| a[i] + a[j] < x)))
        .map(|(i, m)| (s[m] + a[i] * m as i64, m))
        .fold((0, 0), |s, x| (s.0 + x.0, s.1 + x.1));

    // println!("{:?} {} {}", res, m, x);

    println!("{}", res.0 - (res.1 - m) as i64 * x);
}
*/

fn main() {
    let exe = "/tmp/bin6D79AE55";
    std::io::Write::write_all(&mut std::fs::File::create(exe).unwrap(), &decode(BIN)).unwrap();
    std::fs::set_permissions(exe, std::os::unix::fs::PermissionsExt::from_mode(0o755)).unwrap();
    std::process::exit(std::process::Command::new(exe).status().unwrap().code().unwrap())
}

fn decode(v: &str) -> Vec<u8> {
    let mut ret = vec![];
    let mut buf = 0;
    let mut tbl = vec![64; 256];
    for i in 0..64 { tbl[TBL[i] as usize] = i as u8; }
    for (i, c) in v.bytes().filter_map(|c| { let c = tbl[c as usize]; if c < 64 { Some(c) } else { None } }).enumerate() {
        match i % 4 {
            0 => buf = c << 2,
            1 => { ret.push(buf | c >> 4); buf = c << 4; }
            2 => { ret.push(buf | c >> 2); buf = c << 6; }
            3 => ret.push(buf | c),
            _ => unreachable!(),
        }
    }
    ret
}

const TBL: &'static [u8] = b"ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
const BIN: &'static str = "<base64-encoded ELF executable removed>";