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)

use std::io::{stdin, Read};

macro_rules! parse {
    ($it: ident) => {
        $it.next().unwrap().parse().unwrap()
    };
    ($it: ident,2) => {
        (parse!($it), parse!($it))
    };
    ($it: ident,3) => {
        (parse!($it), parse!($it), parse!($it))
    };
    ($it: ident,4) => {
        (parse!($it), parse!($it), parse!($it), parse!($it))
    };
    ($it: ident, $n: expr) => {
        (&mut $it).take($n).map(|e| e.parse().unwrap()).collect();
    };
}

fn binary_search<T: Ord>(v: &Vec<T>, x: T) -> usize {
    let mut ok: isize = -1;
    let mut ng: isize = v.len() as isize;
    while ng - ok > 1 {
        let m = (ok + ng) / 2;
        if v[m as usize] >= x {
            ok = m;
        } else {
            ng = m;
        }
    }
    ng as usize
}

fn solve(s: &str) {
    let mut it = s.split_whitespace();
    let (n, M): (usize, usize) = parse!(it, 2);
    let mut a: Vec<usize> = parse!(it, n);
    a.sort();
    a.reverse();
    let b = a
        .iter()
        .scan(0, |s, &x| {
            *s += x;
            Some(*s)
        })
        .collect::<Vec<_>>();
    let p = |i: usize, x: usize| binary_search(&a, if x > a[i] { x - a[i] } else { 0 });
    let mut ok = 0usize;
    let mut ng = std::usize::MAX / 10;
    while ng - ok > 1 {
        let m = (ok + ng) / 2;
        let mut s = 0;
        for i in 0..n {
            let pi = p(i, m);
            s += pi;
        }
        if s >= M {
            ok = m;
        } else {
            ng = m;
        }
    }

    let mut s = 0;
    let mut ret = 0;
    for i in 0..n {
        let pi = p(i, ok);
        s += pi;
        ret += if pi > 0 { b[pi - 1] } else { 0 } + pi * a[i];
    }
    ret -= (s - M) * ok;
    println!("{}", ret);
}

fn main() {
    let mut s = String::new();
    stdin().read_to_string(&mut s).unwrap();
    solve(&s);
}

#[cfg(test)]
mod tests {
    use super::*;
    #[test]
    fn test_input() {
        let s = "
        5 3
10 14 19 34 33
        ";
        solve(s);
    }
}