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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        long m = sc.nextLong();
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = sc.nextInt();
        }
        Arrays.sort(arr);
        long[] sums = new long[n + 1]; 
        sums[n - 1] = arr[n - 1];
        for (int i = n - 2; i >= 0; i--) {
            sums[i] = arr[i] + sums[i + 1];
        }
        TreeMap<Integer, Integer> map = new TreeMap<>();
        for (int i = 0; i < n; i++) {
            map.put(arr[n - 1 - i], i + 1);
        }
        map.put(Integer.MAX_VALUE, 0);
        int left = 0;
        int right = 200000;
        while (right - left > 1) {
            int mm = (left + right) / 2;
            if (getCount(mm, arr, map) >= m) {
                left = mm;
            } else {
                right = mm;
            }
        }
        long count = 0;
        long total = 0;
        for (int i = 0; i < n; i++) {
            Map.Entry<Integer, Integer> entry = map.ceilingEntry(left - arr[i]);
            count += entry.getValue();
            total += sums[n - entry.getValue()] + (long)(arr[i]) * entry.getValue();
        }
        total -= (count - m) * left;
        System.out.println(total);
    }
    
    static long getCount(int m, int[] arr, TreeMap<Integer, Integer> map) {
        long count = 0;
        for (int i = 0; i < arr.length; i++) {
            Map.Entry<Integer, Integer> entry = map.ceilingEntry(m - arr[i]);
            count += entry.getValue();
        }
        return count;
    }
}