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: Time Limit Exceeded)

class HeapArray
  def initialize
    @heapArr = []
  end

  def push(n)
    i = @heapArr.size

    while i>0
      p = (i-1)/2
      break if @heapArr[p] <= n
      @heapArr[i] = @heapArr[p]
      i=p
    end

    @heapArr[i] = n
  end

  def pop
    min = @heapArr[0]
    n   = @heapArr.pop
    unless @heapArr.empty?
      i   = 0
      while (i*2+1) < @heapArr.size
        if @heapArr[i*2+2].nil?
          buf=i*2+1
        else
          buf=(@heapArr[i*2+1]<@heapArr[i*2+2])?i*2+1:i*2+2
        end
        break if @heapArr[buf] >= n
        @heapArr[i] = @heapArr[buf]
        i=buf
      end
      @heapArr[i]=n
    end
    return min
  end
end


ha = HeapArray.new

io = STDIN
n,m=io.gets.split.map(&:to_i)
a=io.gets.split.map(&:to_i)
a.each{|i|ha.push(-i*2)}
n.times do |i|
  (i+1...n).each do |j|
    ha.push(-(a[i]+a[j]))
    ha.push(-(a[i]+a[j]))
  end
end
puts m.times.map{-ha.pop}.inject(:+)