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: Time Limit Exceeded)

// This code is generated by [cargo-atcoder](https://github.com/tanakh/cargo-atcoder)
// Original source code:
/*
use libprocon::*;

fn main() {
    input! {
        N: usize,
        M: usize,
        A: [i64; N],
    }
    let mut A = A;
    A.sort_by(|a, b| b.cmp(a));

    use std::collections::{BinaryHeap, HashSet};

    #[derive(Debug, Copy, Clone, PartialEq, Eq, PartialOrd)]
    struct Happiness {
        point: i64,
        a: usize,
        b: usize,
    }

    impl Ord for Happiness {
        fn cmp(&self, other: &Happiness) -> std::cmp::Ordering {
            self.point.cmp(&other.point)
        }
    }

    let mut q = BinaryHeap::new();
    q.push(Happiness {
        point: A[0] + A[0],
        a: 0,
        b: 0,
    });

    let mut used = HashSet::new();
    used.insert((0, 0));

    let mut ans = 0;

    for _ in 0..M {
        let val = q.pop().unwrap();
        ans += val.point;

        if !used.contains(&(val.a + 1, val.b)) && val.a + 1 < N {
            used.insert((val.a + 1, val.b));
            q.push(Happiness {
                point: A[val.a + 1] + A[val.b],
                a: val.a + 1,
                b: val.b,
            });
        }

        if !used.contains(&(val.a, val.b + 1)) && val.b + 1 < N {
            used.insert((val.a, val.b + 1));
            q.push(Happiness {
                point: A[val.a] + A[val.b + 1],
                a: val.a,
                b: val.b + 1,
            });
        }
    }
    println!("{}", ans);
}

// this approach seems right, but needs to be less computation
fn this_is_tle() {
    input! {
        N: usize,
        M: usize,
        A: [i64; N],
    }
    let mut A = A;
    A.sort_by(|a, b| b.cmp(a));

    let solve = |c: i64, h: &mut Vec<i64>| {
        let mut cnt = 0;
        for i in 0..N {
            for j in 0..N {
                let s = A[i] + A[j];
                if s < c {
                    break;
                }
                cnt += 1;
                h.push(s);
                dbg!(s);
                if cnt >= M {
                    return true;
                }
            }
        }
        false
    };

    let mut ok: i64 = 0;
    let mut ng: i64 = 1 << 60;
    let mut happiness = Vec::new();
    while ng - ok > 1 {
        let mid = (ng + ok) / 2;
        if solve(mid, &mut happiness) {
            ok = mid;
        } else {
            ng = mid;
        }
    }
    dbg!(ok, ng);
    happiness.clear();
    solve(ok + 1, &mut happiness);
    println!(
        "{}",
        happiness
            .into_iter()
            .chain(vec![ok].into_iter().cycle())
            .take(M)
            .sum::<i64>()
    );
}
*/

fn main() {
    let exe = "/tmp/bin22312902";
    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>";