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: Accepted)

class Array
  def cumsum; s = 0; map{|k| s += k } end
  def fd(n=nil); n = self.max if n.nil?; res = Array.new(n+1,0);self.map{|e| res[e]+=1 };res;end
  # cumlative frequency dist.
  def cfd(n=nil); fd.cumsum; end
end

n, m = gets.to_s.split.map{|t|t.to_i}
a    = gets.to_s.split.map{|t|t.to_i}.sort

sum_min = a[0]  * 2
sum_max = a[-1] * 2

CFD = a.fd.reverse.cumsum.reverse
def number_greater_than(x)
  return CFD[0] if x <= 0
  return 0 if  CFD.size <= x+1
  CFD[x+1]
end

sum_x = (sum_min .. sum_max).bsearch do |sum_x|
  cnt = 0
  a.each do |r|
    cnt += number_greater_than(sum_x-r)
    break if cnt >= m
  end
  cnt < m
end

c = ([0] + a).cumsum
s = c[-1]

ans = 0
a.each do |r|
  pos = a.bsearch_index{|l| sum_x < r + l } || n
  cnt = n - pos
  ans += cnt * r + (s - c[pos])
  m -= cnt
end

ans += m * sum_x

puts ans