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 Array
  def cumsum; s = 0; map{|k| s += k } 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

cnt = nil
sum_x = (sum_min .. sum_max).bsearch do |sum_x|
  cnt = 0
  a.each do |r|
    cnt += n - (a.bsearch_index{|l| sum_x < l + r } || n)
    break if cnt >= m
  end
  cnt < m
end
# p m
# p cnt
    counter = Array.new(10**5+2){ 0 }
cum_counter = Array.new(10**5+2){ 0 }

h = Hash.new(0)
a.each{|e| h[e] += 1 }
# h = a.tally
# h.default = 0
# p a.reverse.cumsum

(10**5).downto(0) do |i|
  counter[i] = h[i] + counter[i+1]
  cum_counter[i] = h[i] * i + cum_counter[i+1]
end
# p a
# p sum_x
sum_all = cum_counter[0]
# sum_all = a.sum
ans = 0
a.each do |r|
  
  if sum_x-r <= 0
    t = sum_all + n * r
  else
    next if sum_x-r >= 10 ** 5 + 1
    t = cum_counter[sum_x-r] + counter[sum_x-r] * r
  end
  
  # p [r, sum_x-r, cum_counter[sum_x-r], counter[sum_x-r], t]
  
  ans += t 
  
end
# puts ans
# puts cnt - m

cnt = 0
a.each do |r|
  
  cnt += n - (a.bsearch_index{|l| sum_x <= l + r } || n)
  # break if cnt >= m
end

ans -= sum_x * (cnt - m)
# puts counter[sum_x]

puts ans


# c = a.cumsum.unshift(0)
# s = c[-1]
# # p s
# ans = 0
# a.each do |r|
#   pos = a.bsearch_index{|l| sum_x < r + l } || n
#   # p ["pos:", pos]
#   cnt = n - pos
#   ans += cnt * r + (s - c[pos])
#   m -= cnt
# end

# ans += m * sum_x

# puts ans