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)

fn read_stdin() -> String {
    use std::io::Read;
    let mut input = String::new();
    std::io::stdin().read_to_string(&mut input).unwrap();
    input
}

fn scan<'a, I: Iterator<Item = &'a str>, T: std::str::FromStr>(tokens: &mut I) -> T {
    tokens.next().unwrap().parse().ok().unwrap()
}

fn rep<T, F: FnMut() -> T>(n: usize, mut f: F) -> Vec<T> {
    (0..n).map(|_| f()).collect()
}

fn main() {
    let input = read_stdin();
    let tokens = &mut input.split_whitespace();
    let n: usize = scan(tokens);
    let m: usize = scan(tokens);
    let a: Vec<usize> = {
        let mut a: Vec<usize> = rep(n, || scan(tokens));
        a.sort();
        a.reverse();
        a
    };
    let mut l = 0;
    let mut u = 200001;
    let mut u_count = 0;
    while l + 1 < u {
        let c = (l + u) / 2;
        let mut count = 0;
        let mut j = 0;
        for i in (0..n).rev() {
            while j < n && a[i] + a[j] >= c {
                j += 1;
            }
            count += j;
        }
        if count >= m {
            l = c;
        } else {
            u = c;
            u_count = count;
        }
    }
    let cumsum = {
        let mut cumsum = vec![0; n + 1];
        for i in 0..n {
            cumsum[i + 1] = cumsum[i] + a[i];
        }
        cumsum
    };
    let mut ans = 0;
    let mut j = 0;
    for i in (0..n).rev() {
        while j < n && a[i] + a[j] >= u {
            j += 1;
        }
        ans += j * a[i] + cumsum[j];
    }
    ans += (m - u_count) * l;
    println!("{}", ans);
}