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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<inttypes.h>
int cmp(const void *x,const void *y){
    int64_t a=*(const int64_t*)x;
    int64_t b=*(const int64_t*)y;
    return a>b ?-1:a<b;
}
int N;
int64_t M,A[200000],A_ruiseki[200000];
int64_t gcount(int64_t* sum,int64_t target){
    int i;
    int64_t count=0;
    if(sum)
        *sum=0;
    for(i=0;i<N;i++){
        int ge=0,less=N-1;
        if(A[i]+A[0]<target)
            count+=0;
        else if(A[i]+A[N-1]>=target){
            count+=N;
            if(sum)
                (*sum)+=A[i]*N+A_ruiseki[N-1];
        }
        else{
            while(ge+1<less){
                int mid=ge+(less-ge)/2;
                if(A[i]+A[mid]>=target)
                    ge=mid;
                else{
                    less=mid;
                }
            }
            count+=ge+1;
            if(sum)
                (*sum)+=A[i]*(ge+1)+A_ruiseki[ge];
        }
    }
    return count;
}
int main(void){
    int i;
    int64_t less,ge=0,nyan,wa;
    if(scanf("%d%"SCNd64,&N,&M)!=2) return 1;
    for(i=0;i<N;i++){
        if(scanf("%"SCNd64,&A[i])!=1) return 1;
    }
    qsort(A,N,sizeof(*A),cmp);
    A_ruiseki[0]=A[0];
    for(i=1;i<N;i++)
        A_ruiseki[i]=A_ruiseki[i-1]+A[i];
    less=A[0]*2+1;
    while(less-1>ge){
        int64_t mid=ge+(less-ge)/2;
        if(gcount(NULL,mid)>=M)
            ge=mid;
        else
            less=mid;
    }
    nyan=gcount(&wa,ge);
    printf("%" PRId64 "\n",wa-(nyan-M)*ge);
    return 0;
}