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::*;
use std::iter::Rev;
use std::ops::Range;
use std::str::FromStr;

fn read_vec2(n: u32) -> Vec<Vec<char>> {
    let mut v2 = Vec::with_capacity(n as usize);
    use std::io::stdin;
    use std::io::BufRead;
    let stdin = stdin();
    let mut stdin = stdin.lock();
    for _ in 0..n {
        let mut s = String::new();
        stdin.read_line(&mut s).ok();
        let v = s.trim().chars().filter(|&c| c != ' ').collect();
        v2.push(v);
    }
    v2
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

#[allow(unused_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };

    ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, [ $t:tt ]) => {
        {
            let len = read_value!($next, usize);
            (0..len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
        }
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}
//usage
//input! {
//  h: i32;
//  v: [i32; n]
//  x: [[i32; n]; m]
//  f: chars <- Vec<char>
//}
// let s: type = read();
trait BinarySearch<T> {
    fn lower_bound(&self, x: T) -> usize;
    fn upper_bound(&self, x: T) -> usize;
}

impl<T: Ord> BinarySearch<T> for Vec<T> {
    fn lower_bound(&self, x: T) -> usize {
        let mut low = 0;
        let mut high = self.len();
        while low != high {
            let mid = (high + low) / 2;
            if self[mid] < x {
                low = mid + 1;
            } else {
                high = mid;
            }
        }
        low
    }

    fn upper_bound(&self, x: T) -> usize {
        let mut low = 0;
        let mut high = self.len();

        while low != high {
            let mid = (high + low) / 2;
            if self[mid] > x {
                high = mid;
            } else {
                low = mid + 1;
            }
        }
        low
    }
}

fn main() {
    // a[i]番目の人と左手を繋いでいるとする
    // a[i]を持っているときに和がX未満になるとき右手の数字はx - a[i]未満の数字である
    // lower_boundでこれを判定する
    // この値をNから引くことで和がX以上になるペアの個数がわかる
    // この値の合計がM未満になればそこが最大の幸福度である
    input! {
        n: i64, mut m: i64,
        mut a: [i64; n]
    }
    a.sort();
    let mut ng: i64 = 0;
    let mut ok: i64 = 100000 * (n + 1);

    //何点を選ぶとM個になるかの二分探索
    while 1 < (ok - ng).abs() {
        let mid = (ok + ng) / 2;
        let check = {
            let mut count = 0;
            for i in 0..n as usize {
                let mut position = a.lower_bound(mid - a[i]);
                count += n as usize - position;
            }
            count < m as usize
        };

        if check {
            ok = mid;
        } else {
            ng = mid;
        }
    }
    // ng -> X - 1, ok -> x

    // upper_boundで左手にa[i]がある場合のX - a[i]の数を判定する
    // Nから求めた数を引いてa[i]を左手に持ったときのX以上の数を数える
    // ansに2行目で求めた数*a[i] + 右手に持つ幸福度の総和を足す
    // Mから2行目で求めた数を引く

    let mut ans = 0;
    let mut count: i64 = 0;

    let mut sum_array = vec![0; (n + 1) as usize];
    for i in 0..n as usize {
        sum_array[(i + 1) as usize] = sum_array[i] + a[i];
    }

    for i in 0..n as usize {
        let mut position: i64 = a.upper_bound(ng - a[i]) as i64;
        ans += (n - position) * a[i] + sum_array[n as usize] - sum_array[(position) as usize];
        m -= (n - position);
    }

    ans += ng * m;
    println!("{}", ans);
}