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 (1.42.0) (judge verdict: Accepted)

#![allow(unused_imports)]
use {
   proconio::{fastout, input, marker::*},
   std::cmp::*,
   std::collections::*,
};

#[fastout]
fn main() {
   input! {
      n: usize,
      m: usize,
      mut a: [usize; n]
   }

   a.sort();
   let mut high = a[n - 1] * 2 + 1;
   let mut low = 0;
   while high - low > 1 {
      let mid = (high + low) / 2;
      let mut k = 0;
      let mut c = 0;
      for i in (0..n).rev() {
         while k < n && a[i] + a[k] < mid {
            k += 1;
         }
         if k == n {
            break;
         }
         c += n - k;
      }
      if c >= m {
         low = mid;
      } else {
         high = mid;
      }
   }

   let mut b = a.clone();
   b.insert(0, 0);
   for i in 0..n {
      b[i + 1] += b[i];
   }
   let mut ans = 0;
   let mut k = 0;
   let mut c = 0;
   for i in (0..n).rev() {
      while k < n && a[i] + a[k] < low {
         k += 1;
      }
      if k == n {
         break;
      }
      c += n - k;
      ans += (n - k) * a[i];
      ans += b[n] - b[k];
   }
   ans -= low * (c - m);
   println!("{}", ans);
}