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: Wrong Answer)

#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
using namespace std;

typedef long long ll;
typedef pair<ll,ll> ii;
typedef tree<ii,null_type,less<ii>,rb_tree_tag,tree_order_statistics_node_update> indexed_set;

int main() {
	cin.tie(0); cout.tie(0); ios_base::sync_with_stdio(0);
	ll n,m; cin>>n>>m;
	vector<ll> a(n);
	for (int i=0; i<n; i++)
		cin>>a[i];
	sort(a.rbegin(),a.rend());
	vector<ll> cum(n+1,0);
	for (int i=1; i<=n; i++)
		cum[i]=cum[i-1]+a[i-1];

	indexed_set is;
	for (int i=0; i<n; i++)
		is.insert({a[i],i});

	ll low=0, high=200007, mid=-1;
	ll cur=-1;
	// largest threshold T such that |ai+aj>=T|>=m
	for (int i=0; i<25; i++) {
		mid=(low+high)/2;
		cur=0;
		for (int j=0; j<n; j++) {
			cur+=n-is.order_of_key({mid-a[j],0});
		}
		if(cur<m) high=mid-1;
		else low=mid;
	}
//	cout<<mid<<" "<<cur<<"\n";
	// figure out |ai+aj>=T|=k. remove the smallest ai+aj k-m times.
	ll sum=0;
	for (int i=0; i<n; i++) {
		ll tmp=n-is.order_of_key({mid-a[i],0});
		sum+=tmp*a[i]+cum[tmp];
	}
	sum-=max(cur-m,0LL)*mid;
	cout<<sum<<"\n";
	return 0;
}