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: Accepted)

import java.util.*;
import java.util.stream.Collectors;
import java.util.stream.Stream;

public class Main {
    private static List<Long> sums;
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        long M = sc.nextLong();
//        int[] A = new int[N];
//        for(int i=0;i<N;i++){A[i]=sc.nextInt();}
        sums = new ArrayList<>(N);
        List<Long> A = Stream.generate(sc::nextLong)
                .limit(N)
                .sorted()
                .collect(Collectors.toList());
        long nowSum = 0;
        for(int i=A.size()-1;i>=0;i--){
            nowSum += A.get(i);
            sums.add(nowSum);
        }

        long l=0;
        long r=10_000_000_000_000_000L;
        while(l<r){
            long m = (l+r)/2;
//            System.out.println(m + ":"+getTotalNumOver(A, m));
            if(getTotalNumOver(A, m) > M){
                l = m+1;
            }else{
                r = m;
            }
        }
        long cnt = getTotalNumOver(A,l);
        final long tgtSum = l;
        long sum = 0;
        sum += (l-1) * (M-cnt);

//        assert(getSumOver(A, 67, 34) == 135);
//        assert(getSumOver(A, 67, 10) == 0);
        //l以上な和をすべて足したもの。
        sum += A.stream().mapToLong(
                p->getSumOver(A, tgtSum, p)
        ).sum();
        System.out.println(sum);
    }

    private static long getSumOver(List<Long> lst, long maxsum, long p){
        long tgt = maxsum-p;
        int l=0;
        int r=lst.size()-1;
        while(l<r){
            int m=(l+r)/2;
            if(tgt > lst.get(m)){
                l = m+1;
            }else{
                r = m;
            }
        }
        if(tgt > lst.get(l)){
            return 0;
        }
        return getSumAft(lst, l) + p * (lst.size()-l);

    }
    private static long getSumAft(List<Long> lst, int ind){
        //returns sum after index inclusive.
        return sums.get(lst.size()-ind-1);
    }

    private static long getTotalNumOver(List<Long> lst, long tgt){
        return lst.stream().mapToLong(
                p -> getCountOver(lst, tgt - p)
        ).sum();
    }

    private static long getCountOver(List<Long> lst, long tgt){
        int l=0;
        int r=lst.size()-1;
        while(l<r){
            int m=(l+r)/2;
            if(tgt > lst.get(m)){ // tgt > lst[m]
                l=m+1;
            }else{
                r=m;
            }
        }
        if(tgt > lst.get(l)){
            return 0;
        }else {
            return lst.size()-l;
        }
    }
}