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 Ruby (2.3.3) (judge verdict: Wrong Answer)

N,M = gets.split(" ").map(&:to_i)
A = gets.chomp.split(" ").map(&:to_i).sort.reverse
@count = Array.new(A[0]+1,0)
A.each do |a|
  @count[a] += 1
end

A[0].times do |i|
  @count[-i-2] += @count[-i-1]
end

def 点以上取れるパターン数 x
  count = 0
  N.times do |i|
    point = x - A[i]
    if point < 0
      count += N 
    else
      count += @count[point] || 0
    end
    #if A[0] >= point
    #  left = 0
    #  right = N
    #  while(left+1<right) do
    #    a = (left + right) >>1
    #    if A[a] >= point 
    #      left = a
    #    else
    #      right = a
    #    end
    #  end
    #  count += left + 1
    #end
  end
  count
end

left = 0
right = A[0]*2 + 1
while(left+1<right) do
  x = (left + right) >>1
  m = 点以上取れるパターン数(x)
  # p [x, m]
  if m >= M
    left = x
  else
    right = x
  end
end
x = left
result = 0
m = 0
N.times do |i|
  break if m >= M
  N.times do |j|
    break if m >= M
    if A[i] + A[j] >= x
      result += A[i] + A[j]
      m += 1
    else
      break
    end
  end
end
# p x
# p [M, m]
puts result