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 ll long long
using namespace std;
ll n,m,a[100005],S[100005],ans;

bool check(ll x) {
	ll tot=0;
	for (int i=n; i>=1; i--) {
		int l=1,r=n,mid,res=-1;
		while (l<=r) {
			mid=(l+r)/2;
			if (a[i]+a[mid]>=x) r=mid-1,res=mid;
				else l=mid+1;
		}
		if (res==-1) continue;
		tot+=n-res+1;
	}
	
	if (tot>=m) return 1;
	return 0;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(NULL),cout.tie(NULL);
	
	cin>>n>>m;
	
	for (int i=1; i<=n; i++) 
		cin>>a[i];
	
	sort(a+1,a+n+1);
	
	for (int i=1; i<=n; i++)
		S[i]=S[i-1]+a[i];
	
	ll l=0,r=2*a[n],mid,res=-1,resu=-1;
	
	while (l<=r) {
		mid=(l+r)/2;
		if (check(mid)) res=mid,l=mid+1;
			else r=mid-1;
	}

	ll tot=0;
	
	for (int i=n; i>=1; i--) {
		l=1,r=n,resu=-1;
		while (l<=r) {
			mid=(l+r)/2;
			if (a[i]+a[mid]>=res) r=mid-1,resu=mid;
				else l=mid+1;
		}
		if (resu==-1) continue;
		tot+=n-resu+1;
		ans+=S[n]-S[resu-1]+a[i]*(n-resu+1);
	}
	
	ans-=(tot-m)*res;
	
	cout<<ans<<endl;
	
	return 0;
}