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>

using namespace std;

#define fi first
#define se second
#define eb emplace_back
#define mt make_tuple
#define all(x) (x).begin(), (x).end() 
#define MOD 1000000007

typedef long long ll;
typedef pair <int, int> ii;
typedef pair <ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef long double ld;

const ll INF=LLONG_MAX;
const int mxn=1e5+5;
int a[mxn];
ll dp[mxn];

int main()
{
    ios_base::sync_with_stdio(0); cin.tie(0);
    //freopen("input.txt","r",stdin); freopen("output.txt","w",stdout);
	int n,m; 
	cin >> n >> m;
	for(int i=0; i<n; i++)cin >> a[i];
	sort(a,a+n);
	int le = 0, ri = 2e5+1;
	ll cnt,happiness=0LL;
	while(le<=ri){
		int mid = (le+ri)/2;
		cnt=0;
		for(int i=0; i<n; i++){
			int left = a[i];
			int right = mid-left;
			auto ptr = lower_bound(a,a+n,right)-a;
			cnt+=n-ptr;
		}
		if(cnt>m){
			le=mid+1;
		}else{
			ri=mid-1;
		}
	}
	le--;
	cnt = 0;
	for(int i=n-1; i>=0; i--){
		dp[i]=dp[i+1]+(ll)a[i];
	}
	for(int i=0; i<n; i++){
		int left = a[i];
		int right = le-left;
		auto ptr = lower_bound(a,a+n,right)-a;
		cnt += n-ptr;
		happiness += dp[ptr]+(n-ptr)*left;
	}
	cnt = cnt-m;
	happiness -= cnt*le;
	cout << happiness << endl;
}
	
// READ & UNDERSTAND
// ll, int overflow, array bounds, memset(0)
// special cases (n=1?), n+1 (1-index)
// do smth instead of nothing & stay organized
// WRITE STUFF DOWN