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: Time Limit Exceeded)

#include <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <numeric>
#include <queue>
#include <set>
using namespace std;
typedef long long int ll;

int main(){
	cin.tie(nullptr);
	ios::sync_with_stdio(false);
	ll n,m; cin >> n >> m;
	vector<ll> a(n);
	for(int i=0;i<n;i++){
		cin >> a[i];
	}
	sort(a.begin(),a.end());
	set<int> s[n];
	ll ans=0;
	priority_queue<pair<ll,pair<int,int>>> q;
	if(m<=1e5){
		q.push(make_pair(a[n-1]+a[n-1],make_pair(n-1,n-1)));
		while(m){
			auto p=q.top(); q.pop();
			ans+=p.first;
			m--;
			int k=p.second.first;
			int l=p.second.second;
			if(k-1>=0&&s[k-1].find(l)==s[k-1].end()){
				s[k-1].insert(l);
				q.push(make_pair(a[k-1]+a[l],make_pair(k-1,l)));
			}
			if(l-1>=0&&s[k].find(l-1)==s[k].end()){
				s[k].insert(l-1);
				q.push(make_pair(a[k]+a[l-1],make_pair(k,l-1)));
			}
		}
	}
	else{
		for(int i=0;i<n;i++){
			ans+=a[i]*n*2;
		}
		q.push(make_pair(-a[0]-a[0],make_pair(0,0)));
		while(m){
			auto p=q.top(); q.pop();
			ans+=p.first;
			m--;
			int k=p.second.first;
			int l=p.second.second;
			if(k+1<n&&s[k+1].find(l)==s[k+1].end()){
				s[k+1].insert(l);
				q.push(make_pair(-a[k+1]-a[l],make_pair(k+1,l)));
			}
			if(l+1<n&&s[k].find(l+1)==s[k].end()){
				s[k].insert(l+1);
				q.push(make_pair(-a[k]-a[l+1],make_pair(k,l+1)));
			}
		}
	}
	cout << ans << endl;
}