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 <iostream>
#include <iomanip>
#include <algorithm>
#include <stdio.h>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <utility>
#include <tuple>
#include <math.h>
#include <set>
#include <map>
using namespace std ;
using ll = long long ;
using ld = long double ;
using vll = vector<ll> ;
using vvll = vector<vll> ;
using vc = vector<char> ;
using vvc = vector<vc> ;
using vb = vector<bool> ;
using vvb = vector<vb> ;
using pll = pair<ll,ll> ;
#define all(v) v.begin(),v.end() 
const ll INF = 1e18 ;
const ll mod = 1e9+7 ;
const double pie = acos(-1); 
vll dx4 = {-1,0,1,0} ;
vll dy4 = {0,-1,0,1} ;
vll dx8 = {-1,0,1,1,1,0,-1,-1} ;
vll dy8 = {-1,-1,-1,0,1,1,1,0} ; 

void fix_cout(){cout << fixed << setprecision(20) << endl ; }
ll gcd(ll a,ll b){if(b==0) return a ; return gcd(b,a%b) ; }
ll lcm(ll a,ll b){return a/gcd(a,b)*b ; }
void chmax(ll &a,ll b){if(a<b) a = b ; }
void chmin(ll &a,ll b){if(a>b) a = b ; }

int solve(){
	ll n,m ;
	cin >> n >> m ;
	vll a(n) ;
	for(auto &i:a) cin >> i ;
	sort(all(a)) ; 
	ll l=0,r=a.at(n-1)*2+1,mid ;
	while(l+1<r){
		mid = (l+r)/2 ;
		ll cnt = 0 ;
		for(auto i:a){
			cnt += a.end()-lower_bound(all(a),mid-i) ;
		}
		if(cnt>=m) l = mid ;
		else r = mid ;
	}
	ll ans = 0 ;
	ll cnt = 0 ;
	vll sum(n+1,0) ;
	for(int i=0;i<n;i++) sum.at(i+1) = sum.at(i)+a.at(i) ;
	for(auto i:a){
		auto itr = lower_bound(all(a),r-i) ;
		ll d = itr-a.begin() ;
		ans += sum.at(n)-sum.at(d)+i*(n-d) ;
		cnt += n-d ;
	}
	ans += (m-cnt)*l ;
	cout << ans << endl ;
}

int main(){
	solve() ;
}