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)

/*
Keep clam  Believe youself
*/
#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<stack>
#include<set>
#include<cmath>
#define Xiaobo main
using namespace std;
const int maxn=2e5+7;
const int mod=1e9+7;
const double eps=1e-15;
const double pi=acos(-1);
const int INF=0x3f3f3f3f;
typedef long long ll;
ll read(){ll c = getchar(),Nig = 1,x = 0;while(!isdigit(c) && c!='-')c = getchar();if(c == '-')Nig = -1,c = getchar();while(isdigit(c))x = ((x<<1) + (x<<3)) + (c^'0'),c = getchar();return Nig*x;}
ll gcd(ll a,ll b){ return b==0?a:gcd(b,a%b);}
//别急 dp找状态 问题看看本身的规律 贪心找方法
ll sum[maxn];
ll a[maxn];
bool cmp(ll x,ll y) {
	return x>y;
}
int Xiaobo()
{
	int n;
	ll m;
	cin>>n>>m;
	for(int i=1;i<=n;i++) {
		scanf("%lld",&a[i]);
	} 
	sort(a+1,a+1+n,cmp);
	for(int i=1;i<=n;i++) {
		sum[i]=sum[i-1]+a[i];
	}
	ll ans=0,k=0;
	for(;ans<m;k++) {
		if(2*n*k-k*k<=m) {
			ans=2*n*k-k*k;
		}
		else break;
 	}
 	//cout<<k<<' '<<ans<<'\n';
	ll ans2=0;
	for(int i=1;i<k;i++) {
		ans2+=((n-i+1)*a[i]+sum[n]-sum[i])*2;
	}
	m-=ans;
	//cout<<m<<"\n";
	for(int i=1,j=k;i<=m;i++,j++) {
		ans2+=a[k]+a[j];
		if(i+1<=m&&k!=j) i++,ans2+=a[k]+a[j];
	}
	cout<<ans2<<'\n';
	return 0;
}