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 (1.42.0) (judge verdict: Accepted)

use proconio::input;
use num::Complex;
use std::f64::consts::PI;

fn main() {
    input! {
        n: usize,
        mut m: isize,
        p: [usize; n],
    }

    let max = *p.iter().max().unwrap();
    let mut a = vec![0; max * 2 + 1];
    for &np in &p {
        a[np] += 1;
    }
    let b = a.clone();

    let v = multiply(a, b);

    let mut ans = 0;
    for i in (0..=max*2).rev() {
        //println!("i {} v[{}] {}",i , i, v[i]);
        if m > v[i] {
            m -= v[i];
            ans += i as isize * v[i];
        } else {
            ans += m * i as isize;
            break;
        }
    }
    println!("{}", ans);
}

fn multiply(g: Vec<usize>, h: Vec<usize>) -> Vec<isize> {
    let n = pow_2_at_least(g.len() + h.len() + 1);

    let mut a = vec![Complex::new(0.0, 0.0); n];
    let mut b = vec![Complex::new(0.0, 0.0); n];

    for i in 0..g.len() {
        a[i] = Complex::new(g[i] as f64, 0.0);
    }
    for i in 0..h.len() {
        b[i] = Complex::new(h[i] as f64, 0.0);
    }

    let gg = dft(a, n);
    let hh = dft(b, n);

    let mut ff = vec![];
    for i in 0..n {
        ff.push(gg[i] * hh[i]);
    }
    let ff = inverse_dft(ff, n);
    let mut res = vec![];
    for &c in &ff {
        res.push((c.re / n as f64).round() as isize);
    }
    res
}

fn dft(mut f: Vec<Complex<f64>>, n: usize) -> Vec<Complex<f64>> {
    if n == 1 {
        return f;
    }
    let mut f0 = vec![];
    let mut f1 = vec![];
    for i in 0..n / 2 {
        f0.push(f[2 * i + 0]);
        f1.push(f[2 * i + 1]);
    }
    f0 = dft(f0, n / 2);
    f1 = dft(f1, n / 2);
    let zeta = Complex::new((2.0 * PI / n as f64).cos(), (2.0 * PI / n as f64).sin());
    let mut pow_zeta = Complex::new(1.0, 0.0);

    for i in 0..n {
        f[i] = f0[i % (n / 2)] + pow_zeta * f1[i % (n / 2)];
        pow_zeta *= zeta;
    }
    f
}

fn inverse_dft(f: Vec<Complex<f64>>, n: usize) -> Vec<Complex<f64>> {
    if n == 1 {
        return f;
    }
    let mut f0 = vec![];
    let mut f1 = vec![];
    for i in 0..n / 2 {
        f0.push(f[2 * i + 0]);
        f1.push(f[2 * i + 1]);
    }
    f0 = inverse_dft(f0, n / 2);
    f1 = inverse_dft(f1, n / 2);
    let zeta = Complex::new((2.0 * PI / n as f64).cos(), (2.0 * PI / n as f64).sin());
    let zeta = zeta.inv();
    let mut pow_zeta = Complex::new(1.0, 0.0);

    let mut res = vec![];
    for i in 0..n {
        res.push(f0[i % (n / 2)] + pow_zeta * f1[i % (n / 2)]);
        pow_zeta *= zeta;
    }
    res
}


fn pow_2_at_least(n: usize) -> usize {
    let mut sz = 1;
    while sz <= n {
        sz *= 2;
    }
    sz
}