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: Wrong Answer)

#include<stdio.h>
int partition(int arr[], int low, int high){
    int key;
    key = arr[low];
    while(low<high){
        while(low <high && arr[high]>= key )
            high--;
        if(low<high)
            arr[low++] = arr[high];
        while( low<high && arr[low]<=key )
            low++;
        if(low<high)
            arr[high--] = arr[low];
    }
    arr[low] = key;
    return low;
}
void quick_sort(int arr[], int start, int end){
    int pos;
    if (start<end){
        pos = partition(arr, start, end);
        quick_sort(arr,start,pos-1);
        quick_sort(arr,pos+1,end);
    }
    return;
}

int main()
{
   int n,m;
     int a[100005];
    int b[100005];
    int hash[100005]={0};
    //printf("%d",hash[1231]);
    int ans=0;
    scanf("%d %d",&n,&m);
    for( int i =0;i<n;i++)
    {
        scanf("%d",&a[i]);
        if(hash[a[i]]==0)
            hash[a[i]]++;
    }
    int counter = 0;
    for(int i = 100005;i>=1;i--)
    {
        if(hash[i]==1)
        {
            hash[i]--;
            b[counter]=i;
            counter++;
        }
    }
    //for(int i = 0;i<counter;i++)
     //   printf("%d,",b[i]);
    //sort();
    //quick_sort(a,0,n-1);
    int timer=0;
    for(int i = 0;i<counter;i++)
    {
        for(int j = i;j<counter;j++)
        {
            if(i==j)
            {
                ans+=2*b[i];
                timer++;
                if(timer>=m)
                    goto end;
            }
            else
            {
                ans+=b[i]+b[j];
                timer++;
                if(timer>=m)
                    goto end;
                ans+=b[j]+b[i];
                timer++;
                if(timer>=m)
                    goto end;
            }
        }
    }
    end:;
    printf("%d",ans);
    return 0;
}