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)

$n, $m = gets.split.map &:to_i
$as = gets.split.map &:to_i

$as.sort!

$as_expand = Array.new(10**5 + 1, 0)
$as.each{ |a|
	$as_expand[a] += 1
}

$sum_tail = Array.new(10**5 * 2 + 1, 0)
$count_tail = Array.new(10**5 * 2 + 1, 0)

(10**5).downto(0){ |i|
	if $as_expand[i] != 0 then
		$count_tail[i] = $count_tail[i + 1] + $as_expand[i]
		$sum_tail[i] = $sum_tail[i + 1] + i * $as_expand[i]
	else
		$count_tail[i] = $count_tail[i + 1] 
		$sum_tail[i] = $sum_tail[i + 1]
	end
}

# 握手による幸福度向上値をxとし、(x以上の組み合わせの数)>=mとなるxを2項探索で探す

def calc(x, is_pruned)
	count = 0

	$as.each_with_index{ |a,i|
		if x > a then
			count += $count_tail[x - a]
		else
			count += $n
		end
		
		# 探索中はM以上になったら即打ち切り
		if is_pruned && count >= $m then
			break
		end
	}

	count
end

def calc_sum(x)
	sum = 0

	$as.each_with_index{ |a, i|
		if x > a then
			sum += $sum_tail[x - a] + a * $count_tail[x - a]
		else
			sum += $sum_tail[0] + a * $n
		end
	}

	sum
end


min = $as[0] * 2
max = $as[-1] * 2
current = (min + max) / 2
result = nil

while min != max 
	result = calc(current, true)
	
	if result >= $m then
		if current != min then
			min = current
		else
			break
		end
	else
		max = current
	end

	current = (min + max) / 2
end

# 取り直し
result = calc(current, false)

score = calc_sum(current)

if result != $m then
	score -= current * (result - $m)
end

puts score