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 Fortran (gfortran v4.8.4) (judge verdict: Runtime Error)

    module m_sort
    !reference
    !https://qiita.com/cure_honey/items/29944920e8aff5651e41
      implicit none
    contains  
      recursive function iqsort(ix) result(ires)
        integer(16), allocatable :: ires(:)
        integer(16), intent(in)  :: ix(:)
        integer(16) :: k
        if (size(ix) <= 1) then 
          ires = ix
        else  
          k = ix(size(ix) / 2)
          ires = [iqsort(pack(ix, ix < k)), pack(ix, ix == k), iqsort(pack(ix, ix > k))] 
        end if
      end function iqsort  
    end module m_sort
      
      PROGRAM handshake
      USE m_sort
      IMPLICIT NONE
      integer(16) :: n,m
      integer(16),allocatable :: a(:),sumPow(:)
      integer(16) :: i
      
      read*,n,m
      allocate( a(n),sumPow(n*n) )
      read*,a(:)
      
      do i = 1,n
        sumPow((i-1)*n+1:i*n) = a(i) + a(1:n)
      end do
      sumPow = iqsort( sumPow )
      
      print*,sum(sumPow(n*n-m+1:n*n))
      
      END PROGRAM