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.*;
import java.io.*;
 
public class Main {
    static boolean[][] used;
    static int[] ans;
    public static void main(String[] args) throws Exception {
        long max = 100000 * 20000000000L;
        FastScanner sc = new FastScanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int n = sc.nextInt();
        long m = sc.nextLong();
        int[] a = sc.nextIntArray(n);
        long[] cumA = new long[n+1];
        int[] cumN = new int[200001];
        Arrays.sort(a);
        for(int i = 0; i < n; i++){
            cumA[i+1] = a[i] + cumA[i];
            cumN[a[i]]++;
        }
        for(int i = cumN.length-1; i > 0; i--){
            cumN[i-1] += cumN[i];
        }
        int left = -1;
        int right = 200000+1;
        while(right - left > 1){
            int mid = (left+right)/2;
            long cnt = 0;
            for(int v : a){
                cnt += cumN[Math.max(0,mid-v)];
            }
            if(cnt >= m){
                left = mid;
            }else{
                right = mid;
            }
        }
        long ans = 0;
        int total = 0;
        for(int v : a){
            int cnt = cumN[Math.max(0,left-v)];
            total += cnt;
            ans += v*cnt+(cumA[n]-cumA[n-cnt]);
        }
        System.out.println(ans-(total-m)*left);
    }
}

class FastScanner {
    private BufferedReader reader = null;
    private StringTokenizer tokenizer = null;
    public FastScanner(InputStream in) {
        reader = new BufferedReader(new InputStreamReader(in));
        tokenizer = null;
    }

    public String next() {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken();
    }

    public String nextLine() {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                return reader.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken("\n");
    }

    public long nextLong() {
        return Long.parseLong(next());
    }

    public int nextInt() {
        return Integer.parseInt(next());
    }

    public double nextDouble() {
         return Double.parseDouble(next());
    }

    public int[] nextIntArray(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++)
            a[i] = nextInt();
        return a;
    }

    public long[] nextLongArray(int n) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++)
            a[i] = nextLong();
        return a;
    } 
}