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: Wrong Answer)

#![allow(unused_imports, unused_macros, dead_code)]

macro_rules! min {
    (.. $x:expr) => {{
        let mut it = $x.iter();
        it.next().map(|z| it.fold(z, |x, y| min!(x, y)))
    }};
    ($x:expr) => ($x);
    ($x:expr, $($ys:expr),*) => {{
        let t = min!($($ys),*);
        if $x < t { $x } else { t }
    }}
}
macro_rules! max {
    (.. $x:expr) => {{
        let mut it = $x.iter();
        it.next().map(|z| it.fold(z, |x, y| max!(x, y)))
    }};
    ($x:expr) => ($x);
    ($x:expr, $($ys:expr),*) => {{
        let t = max!($($ys),*);
        if $x > t { $x } else { t }
    }}
}
macro_rules! ewriteln {
    ($($args:expr),*) => { let _ = writeln!(&mut std::io::stderr(), $($args),*); };
}
macro_rules! trace {
    ($x:expr) => { ewriteln!(">>> {} = {:?}", stringify!($x), $x) };
    ($($xs:expr),*) => { trace!(($($xs),*)) }
}
macro_rules! put {
    (.. $x:expr) => {{
        let mut it = $x.iter();
        if let Some(x) = it.next() { print!("{}", x); }
        for x in it { print!(" {}", x); }
        println!("");
    }};
    ($x:expr) => { println!("{}", $x) };
    ($x:expr, $($xs:expr),*) => { print!("{} ", $x); put!($($xs),*) }
}

const M: i64 = 1_000_000_007;

fn bicount(xs: &Vec<i64>, acc: &Vec<i64>, base: i64, z: i64) -> (usize, i64) {
    // trace!("bincount", &xs, &acc, base, z);
    let n = xs.len();
    if xs[n-1] + base >= z { return (n, acc[n]) }
    if xs[0] + base < z { return (0, 0) }

    let mut left = 0;  // base + xs[left] >= z
    let mut right = n;
    for _ in 0..100 {
        let mid = (left + right) / 2;
        if xs[mid] + base >= z {
            left = mid;
        } else {
            right = mid;
        }
    }
    (left + 1, acc[left + 1])
}

fn main() {
    let mut sc = Scanner::new();
    let n: usize = sc.cin();
    let m: usize = sc.cin();
    let mut xs: Vec<i64> = sc.vec(n);
    xs.sort();
    xs.reverse();

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

    if m == 1 { put!(xs[n-1] * 2); return }
    if m == 2 { put!(xs[n-1] * 2 + xs[n-1] + xs[n-2]); return }
    // if m == 3 { put!(xs[n-1] * 2 + (xs[n-1] + xs[n-2]) * 2); return }

    let mut left = 0;  // left 以上は M 個以上ある
    let mut right = xs[0] * 2;  // right 以上は M もない
    for _ in 0..100 {
        let mid = (left + right) / 2;
        let mut count = 0;
        for &x in xs.iter() {
            let (c, _) = bicount(&xs, &acc, x, mid);  // (x + _) が mid 以上の個数と和
            count += c;
        }
        if count >= m {
            left = mid;
        } else {
            right = mid;
        }
    }
    // trace!(&xs);
    // trace!(&acc);
    // trace!(left);
    {
        let mut ans = 0;
        let mut count = 0;
        for &x in xs.iter() {
            let (c, sum) = bicount(&xs, &acc, x, left);
            count += c;
            ans += sum + c as i64 * x;
            // trace!( sum + c as i64 * x );
            // trace!(x, (c, sum));
        }
        ans -= (count - m) as i64 * left as i64;
        put!(ans);
    }

}

use std::io::{self, Write};
use std::str::FromStr;
use std::collections::VecDeque;

struct Scanner { stdin: io::Stdin, buffer: VecDeque<String>, }
impl Scanner {
    fn new() -> Self { Scanner { stdin: io::stdin(), buffer: VecDeque::new() } }
    fn cin<T: FromStr>(&mut self) -> T {
        while self.buffer.len() == 0 {
            let mut line = String::new();
            let _ = self.stdin.read_line(&mut line);
            for w in line.split_whitespace() {
                self.buffer.push_back(String::from(w));
            }
        }
        self.buffer.pop_front().unwrap().parse::<T>().ok().unwrap()
    }
    fn chars(&mut self) -> Vec<char> { self.cin::<String>().chars().collect() }
    fn vec<T: FromStr>(&mut self, n: usize) -> Vec<T> { (0..n).map(|_| self.cin()).collect() }
}