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)

#[allow(unused_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };

    ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

use std::cmp::min;

fn main() {
    input! {
        n: usize,
        mut mm: usize,
        mut a: [i64; n],
    }

    a.sort();
    a.reverse();

    let mut psum = vec![0; n + 1];

    for i in 0..n {
        psum[i + 1] += psum[i] + a[i];
    }

    let mut l = 0_i64;
    let mut r = 200000 + 1;

    while (r - l) > 1 {
        // dbg!(l, r);
        let m = (l + r) / 2;

        let mut cnt = 0;
        for i in 0..n {
            let t = m - a[i];

            let mut ll = i;
            let mut rr = n;
            while (rr - ll) > 1 {
                let mm = (ll + rr) / 2;

                if a[mm] >= t {
                    ll = mm;
                } else {
                    rr = mm;
                }
            }

            // eprintln!("{}: {} {} {}", i, t, ll, rr);

            if a[i] >= t {
                cnt += 1;
                cnt += (ll - i) * 2;
            }
        }

        // dbg!("### {}: {}", m, cnt);

        if cnt >= mm {
            l = m;
        } else {
            r = m;
        }
    }

    // dbg!(l);

    let mut cnt = 0;
    let mut sum = 0;
    for i in 0..n {
        let t = (l + 1) - a[i];

        let mut ll = i;
        let mut rr = n;
        while (rr - ll) > 1 {
            let mm = (ll + rr) / 2;

            if a[mm] >= t {
                ll = mm;
            } else {
                rr = mm;
            }
        }

        if a[i] >= t {
            cnt += 1;
            cnt += (ll - i) * 2;

            sum += a[i] + a[i];
            sum += (psum[ll + 1] - psum[i + 1]) * 2 + a[i] * (ll - i) as i64 * 2;
        }
    }

    // dbg!(sum, l, mm, cnt);

    sum += l * (mm - cnt) as i64;

    println!("{}", sum);
}