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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

#define pb push_back
#define ss second
#define ff first
#define N 100005
#define inf 1000000009
#define ll long long
#define mid(a,b) (a+b)/2
#define pii pair<int,int>

using namespace std;

int n,a[N],s[2 * N],l,r,md,cep,sag,x;
ll ans,jog,den;
ll dg,p[N],sum,m;

int main()
{
	scanf("%d%lld",&n,&m);
	for(int i = 1;i <= n;i++){
		scanf("%d",a + i);
		s[a[i]]++;
	}
	sort(a + 1,a + n + 1);
	for(int i = 1;i <= n;i++)
		p[i] = p[i - 1] + a[i];
	l = 1,r = 2 * 1e5;
	while(l <= r){
		md = mid(l,r);
		ans = den = sum = 0;
		for(int i = 1;i <= n;i++){
			cep = 1,sag = n;
			jog = n + 1;
			while(cep <= sag){
				x = mid(cep,sag);
				if(a[x] + a[i] >= md){
					jog = x;
					sag = x - 1;
				}
				else cep = x + 1;
			}
			if(jog == n + 1) continue;
			ans += n - jog + 1;
			sum += p[n] - p[jog - 1] + (n - jog + 1) * a[i];
			if(a[i] + a[i] == md) den += s[a[i]] - 1;
			else if(a[i] < md) den += s[md - a[i]];
			
		}
		
		if(ans >= m){
			ans -= den;
			sum -= den * md;
			if(ans <= m) dg = sum + (m - ans) * md;
			l = md + 1;
		}	
		else r = md - 1;
	}
	printf("%lld",dg);
}
/*
9 14
1 3 5 110 24 21 34 5 3
*/