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)

def max(a,b); a > b ? a : b; end
  
A_MAX = 100000

N, M = gets.split.map(&:to_i)
A = gets.split.map(&:to_i)
C = Array.new(A_MAX + 10, 0) # C[x] := A.count{|a| a >= x }; C[x] - C[x + 1] := A.count{|a| a == x }
S = Array.new(A_MAX + 10, 0) # S[x] := A.select{|a| a >= x }.sum

A.each{|a| C[a] += 1 }
A_MAX.downto(0) do |i| 
  S[i] = C[i] * i + S[i + 1]
  C[i] += C[i + 1]
end

ANS = {}

xy_ = (1 .. 2 * A_MAX).bsearch do |xy|
  m = 0
  s = 0
  A.each do |x|
    next if xy - x > A_MAX
    m += C[max(xy - x, 0)]
    s += S[max(xy - x, 0)] + C[max(xy - x, 0)] * x
  end
  ANS[xy] = s + (M - m) * (xy - 1)
  m <= M
end
puts ANS[xy_]