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)

n,m=gets.chomp.split.map(&:to_i)
a = []
a = gets.chomp.split.map(&:to_i)
a = a.sort
x = y = n - 1
sum = a[x] * 2
m = m - 1
temp = []
while m > 0
  if y == 0
  	y = x - 1
  	x = x - 1
  else
    y = y - 1
  end
  if x > 0 && a[x] + a[y] < a[x - 1] + a[x - 1]
  	while y >= 0
  	  temp << a[x] + a[y]
  	  y = y - 1
  	end
  	y = x - 1
  	x = x - 1
  	temp = temp.sort
  end
  while temp.count > 0 && m > 0 && temp[temp.count - 1] > a[x] + a[y]
	if m > 1
	  sum = sum + temp.pop() * 2
	  m = m - 2
	else
	  sum = sum + temp.pop()
	  m = m - 1
	end
  end
  break if m == 0
  if x == y || m == 1
	sum = sum + a[x] + a[y]
	m = m - 1
  else
	sum = sum + (a[x] + a[y]) * 2
	m = m - 2
  end
end
print("#{sum}\n")