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<stdio.h>
int selectMaxMark(int* a,int len){
	int max=0;
	int mark=0;
	int i;
	for(i=0;i<len;i++){
		if(a[i]>=max) {
			max=a[i];
			mark=i;
		}
	}
	return mark;
}
void Sort(int a[], int len)
{
    int i;
    int j;
    double temp;  
    int gap;  //步长
    for (gap=len/2;gap>=1;gap/=2) {  // 步长初始化为数组长度的一半，每次遍历后步长减半
        for (i=0+gap;i<len;i+=gap) {   //对步长为gap的元素进行直插排序，当gap为1时，就是直插排序
            temp=a[i];  //备份a[i]的值
            
            j=i-gap; 
            while (j>=0&&a[j]>temp) {
                a[j+gap]=a[j];  
                
                j-=gap;
            }
            a[j+gap] = temp;  
			
        }
    }
}

int main(){
	int N,M;
	scanf("%d %d",&N,&M);
	long long total=0;
	int score[N];
	int i;
	for(i=0;i<N;i++) scanf("%d",&score[i]);
	Sort(score,N);
	int NowMax[N];
	for(i=0;i<N;i++) NowMax[i]=score[i]+score[N-1];
	
	int slc[N];
	for(i=0;i<N;i++) slc[i]=1;
	
	int loop;
	
	for(loop=0;loop<M;loop++){
	
	
	int mark;
	mark=selectMaxMark(NowMax,N);
	total+=NowMax[mark];
	slc[mark]++;
	NowMax[mark]=score[mark]+score[N-slc[mark]];
	if(N-slc[mark]<0) NowMax[mark]=0;
	
	
}
	printf("%lld",total);
}