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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Main main = new Main();
        main.solve();
    }
    public void solve() {
        Scanner scan = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        int N = scan.nextInt();
        long M = scan.nextLong();
        List<Integer> A = new ArrayList<>();
        for (int i = 0; i < N; i++) {
            A.add(scan.nextInt());
        }
        A.sort(Comparator.reverseOrder());
        long[] sumA = new long[N+1];
        for (int i = 0; i < N; i++) {
            sumA[i+1] = sumA[i] + A.get(i);
        }

        long left = 0;
        long right = 20000000000L;
        while (left + 1 != right) {
            long mid = (left + right) / 2;
            long count = count(N, mid, A);
            if (M <= count) {
                left = mid;
            } else {
                right = mid;
            }
        }
        long ans = sum(N, M, left, A, sumA);
        System.out.println(ans);
    }

    private long sum(int N, long M, long threshold, List<Integer> A, long[] sumA) {
        long sum = 0;
        for (int i = 0; i < N && 0 < M; i++) {
            int left = 0;
            int right = N - 1;
            while (left + 1 != right) {
                int mid = (left + right) / 2;
                if (threshold - A.get(i) <= A.get(mid)) {
                    left = mid;
                } else {
                    right = mid;
                }
            }
            int count = (int) Math.min(M,left + 1);
            M -= count;
            sum += (long) A.get(i) * (long) count;
            sum += sumA[count];
        }
        return sum;
    }
    private long count(int N, long threshold, List<Integer> A) {
        long count = 0;
        for (int i = 0; i < N; i++) {
            int left = 0;
            int right = N - 1;
            while (left + 1 != right) {
                int mid = (left + right) / 2;
                if (threshold - A.get(i) <= A.get(mid)) {
                    left = mid;
                } else {
                    right = mid;
                }
            }
            count += left + 1;
        }
        return count;
    }
}