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_imports)]
use std::cmp::Ordering;

#[allow(unused_imports)]
use std::collections::{BTreeMap, BTreeSet, BinaryHeap, HashMap, HashSet, VecDeque};

#[allow(unused_imports)]
use std::io::{stdin, stdout, BufWriter, Write};

#[allow(unused_imports)]
use std::iter::FromIterator;

#[allow(unused_macros)]
macro_rules! min {
    ($x: expr) => ($x);
    ($x: expr, $($z: expr),+) => {{
        std::cmp::min($x, min!($($z),*))
    }}
}

#[allow(unused_macros)]
macro_rules! max {
    ($x: expr) => ($x);
    ($x: expr, $($z: expr),+) => {{
        std::cmp::max($x, max!($($z),*))
    }}
}

#[allow(unused_macros)]
macro_rules! get {
    ([$t: ty]) => {
        {
            let mut line = String::new();
            stdin().read_line(&mut line).unwrap();
            line.split_whitespace().map(|t|t.parse::<$t>().unwrap()).collect::<Vec<_>>()
        }
    };
    ([$t: ty];$n: expr) => {
        (0..$n).map(|_|get!([$t])).collect::<Vec<_>>()
    };
    ($t: ty) => {
        {
            let mut line = String::new();
            stdin().read_line(&mut line).unwrap();
            line.trim().parse::<$t>().unwrap()
        }
    };
    ($($t: ty),*) => {
        {
            let mut line = String::new();
            stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            ($(iter.next().unwrap().parse::<$t>().unwrap(),)*)
        }
    };
    ($t: ty; $n: expr) => {
        (0..$n).map(|_|get!($t)).collect::<Vec<_>>()
    };
    ($($t: ty),*; $n: expr) => {
        (0..$n).map(|_|get!($($t),*)).collect::<Vec<_>>()
    };
}

#[allow(unused_macros)]
macro_rules! debug {
    ($($a:expr),*) => {
        #[cfg(debug_assertions)]
        writeln!(&mut std::io::stderr(), concat!("[DEBUG] ", $(stringify!($a), "={:?} "),*), $($a),*).unwrap();
    }
}

const BIG_STACK_SIZE: bool = true;

#[allow(dead_code)]
fn main() {
    use std::thread;
    if BIG_STACK_SIZE {
        thread::Builder::new()
            .stack_size(32 * 1024 * 1024)
            .name("solve".into())
            .spawn(solve)
            .unwrap()
            .join()
            .unwrap();
    } else {
        solve();
    }
}

fn lower_bound<F>(l: isize, r: isize, criterion: F) -> isize
where
    F: Fn(isize) -> bool,
{
    let mut l = l;
    let mut r = r;
    while r - l > 1 {
        let mid = l + (r - l) / 2;
        if criterion(mid) {
            r = mid;
        } else {
            l = mid;
        }
    }
    l
}

fn solve() {
    let (n, m) = get!(usize, usize);
    let mut a = get!([usize]);

    a.sort();

    let mut sums = vec![0; n + 1];
    for i in 0..n {
        sums[i + 1] = sums[i] + a[i];
    }

    let th = n * n - m;
    let crit = |val: isize| {
        let val = val as usize;
        let mut sum = 0;
        for i in 0..n {
            let crit = |j: isize| {
                let j = j as usize;
                a[i] + a[j] >= val
            };
            let j = (lower_bound(-1, n as isize, crit) + 1) as usize;
            sum += j;
        }
        sum >= th
    };
    let val = lower_bound(0, 1_000_000_000_000_010, crit) as usize;
    let mut res = 0;
    let mut sum = 0;
    for i in 0..n {
        let crit = |j: isize| {
            let j = j as usize;
            a[i] + a[j] >= val
        };
        let j = (lower_bound(-1, n as isize, crit) + 1) as usize;
        sum += j;
        if i >= j {
            continue;
        }
        res += 2 * (a[i] * (j - i) + sums[j] - sums[i]) - 2 * a[i];
    }

    println!("{}", 2 * n * sums[n] - res - (th - sum) * val);
}