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<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
#include<queue>
#include<string>
#include<iostream>

#define maxn 200005
#define INF 0x3f3f3f3f

using namespace std;

inline long long getint()
{
    long long num=0,flag=1;char c;
    while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;
    while(c>='0'&&c<='9')num=num*10+c-48,c=getchar();
    return num*flag;
}

int n;
long long m;
long long a[maxn];
long long sum[maxn];
long long ans;
inline bool cmp(long long a,long long b){return a>b;}

inline bool check(int mid)
{
	int now=n;
	long long num=0;
	for(int i=1;i<=n;i++)
	{
		while(a[i]+a[now]<mid&&now)now--;
		num+=now;
	}
	return num>=m;
}

inline void solve(int mid)
{
	long long mn=INF;
	int now=n;
	long long num=0;
	for(int i=1;i<=n;i++)
	{
		while(a[i]+a[now]<mid&&now)now--;
		ans+=a[i]*now+sum[now];
		mn=min(mn,a[i]+a[now]);
		num+=now;
	}
	printf("%lld\n",ans-(num-m)*mn);
}

int main()
{
	n=getint(),m=getint();
	for(int i=1;i<=n;i++)a[i]=getint();
	sort(a+1,a+n+1,cmp);
	for(int i=1;i<=n;i++)sum[i]=a[i]+sum[i-1];
	int l=0,r=INF;
	while(l<r)
	{
		int mid=(l+r)>>1;if(mid==l)mid++;
		if(check(mid))l=mid;
		else r=mid-1;
	}
	solve(l);
}