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: Runtime Error)

import java.util.*;

class Main {
    static int N;
    static int[] A, B;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        N = 100005;
        int n = sc.nextInt();
        int M = sc.nextInt();
        A = new int[N+1];
        B = new int[N+1];
        for (int i = 0; i < n; ++i) {
            int X = sc.nextInt();
            A[X]++;
            B[X]++;
        }
        ArrayList<Long> list = mul(B, B);
        // for (int i=0;i<15;i++) {
        //     System.out.println(list.get(i));
        // }

        int tmp_M = M;
        long ans = 0;
        for (int i=list.size()-1;i>=0;i--) {
            if (list.get(i)<tmp_M) {
                tmp_M-=list.get(i);
                ans += list.get(i)*(i+1);
            } else {
                ans += tmp_M*(i+1);
                tmp_M=0;
            }
            if (tmp_M<=0) break;
        }
        System.out.println(ans);
    }

    static ArrayList<Long> mul(int[] a, int[] b) {
        int n = 1;
        while (n < a.length + b.length)
            n *= 2;
        Complex[] ac = new Complex[n];
        Complex[] bc = new Complex[n];
        for (int i = 0; i < n; ++i) {
            ac[i] = new Complex(0, 0);
            bc[i] = new Complex(0, 0);
        }
        for (int i = 0; i < a.length; ++i) {
            ac[i].re = a[i];
        }
        for (int i = 0; i < b.length; ++i) {
            bc[i].re = b[i];
        }
        ac = fft(ac, false);
        bc = fft(bc, false);
        for (int i = 0; i < ac.length; ++i) {
            ac[i] = ac[i].mul(bc[i]);
        }
        ac = fft(ac, true);
        for (int i = 0; i < ac.length; ++i) {
            ac[i].re /= n;
            ac[i].co /= n;
        }
        ArrayList<Long> list = new ArrayList<Long>();
        for (int i = 1; i <= 2 * N; ++i) {
            // System.out.println(Math.round(ac[i].re));
            list.add((long)Math.round(ac[i].re));
        }
        return list;

    }

    static Complex[] fft(Complex[] a, boolean rev) {
        int n = a.length;
        if (n == 1)
            return a;
        int c = 0;
        for (int i = 1; i < n; ++i) {
            int j;
            for (j = n >> 1; j > (c ^= j); j >>= 1)
                ;
            if (c > i) {
                Complex tmp = a[c];
                a[c] = a[i];
                a[i] = tmp;
            }
        }

        for (int d = 1; d < n; d <<= 1) {
            for (int j = 0; j < d; ++j) {
                Complex mul = exp(2 * Math.PI / (2 * d) * (rev ? -1 : 1) * j);
                for (int pos = 0; pos < n; pos += 2 * d) {
                    double ure = a[pos + j].re;
                    double uco = a[pos + j].co;
                    double vre = a[pos + j + d].re * mul.re - a[pos + j + d].co * mul.co;
                    double vco = a[pos + j + d].co * mul.re + a[pos + j + d].re * mul.co;
                    a[pos + j].re = ure + vre;
                    a[pos + j].co = uco + vco;
                    a[pos + j + d].re = ure - vre;
                    a[pos + j + d].co = uco - vco;
                }
            }
        }
        return a;
    }

    static class Complex {
        double re, co;

        public Complex(double re, double co) {
            this.re = re;
            this.co = co;
        }

        Complex add(Complex o) {
            return new Complex(re + o.re, co + o.co);
        }

        Complex subtract(Complex o) {
            return new Complex(re - o.re, co - o.co);
        }

        Complex mul(Complex o) {
            return new Complex(re * o.re - co * o.co, re * o.co + o.re * co);
        }
    }

    static Complex exp(double theta) {
        return new Complex(Math.cos(theta), Math.sin(theta));
    }
}