Problem:
Score : 300 points

Problem Statement

You have N bamboos. The lengths (in centimeters) of these are l_1, l_2, ..., l_N, respectively.

Your objective is to use some of these bamboos (possibly all) to obtain three bamboos of length A, B, C. For that, you can use the following three kinds of magics any number:

Extension Magic: Consumes 1 MP (magic point). Choose one bamboo and increase its length by 1.

Shortening Magic: Consumes 1 MP. Choose one bamboo of length at least 2 and decrease its length by 1.

Composition Magic: Consumes 10 MP. Choose two bamboos and combine them into one bamboo. The length of this new bamboo is equal to the sum of the lengths of the two bamboos combined. (Afterwards, further magics can be used on this bamboo.)

At least how much MP is needed to achieve the objective?

Constraints

3 \leq N \leq 8

1 \leq C < B < A \leq 1000

1 \leq l_i \leq 1000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N A B C
l_1
l_2
:
l_N

Output

Print the minimum amount of MP needed to achieve the objective.

Sample Input 1

5 100 90 80
98
40
30
21
80

Sample Output 1

23

We are obtaining three bamboos of lengths 100, 90, 80 from five bamboos 98, 40, 30, 21, 80. We already have a bamboo of length 80, and we can obtain bamboos of lengths 100, 90 by using the magics as follows at the total cost of 23 MP, which is optimal.

Use Extension Magic twice on the bamboo of length 98 to obtain a bamboo of length 100. (MP consumed: 2)

Use Composition Magic on the bamboos of lengths 40, 30 to obtain a bamboo of length 70. (MP consumed: 10)

Use Shortening Magic once on the bamboo of length 21 to obtain a bamboo of length 20. (MP consumed: 1)

Use Composition Magic on the bamboo of length 70 obtained in step 2 and the bamboo of length 20 obtained in step 3 to obtain a bamboo of length 90. (MP consumed: 10)

Sample Input 2

8 100 90 80
100
100
90
90
90
80
80
80

Sample Output 2

0

If we already have all bamboos of the desired lengths, the amount of MP needed is 0. As seen here, we do not necessarily need to use all the bamboos.

Sample Input 3

8 1000 800 100
300
333
400
444
500
555
600
666

Sample Output 3

243

Solution in Ruby (2.3.3) (judge verdict: Runtime Error)

debug = 1

def count_mp(needs, have, join)
  mp = 0
  3.times.each do |i|
    mp += (needs[i]-have[i]).abs
  end
  mp + join * 10
end

N, A, B, C = gets.split.map(&:to_i)
needs = [A, B, C].sort # [80, 90, 100]
p needs if debug == 1
bamboos = N.times.map { gets.to_i }
bamboos.sort! # [21, 30, 40, 80, 98]
p bamboos if debug == 1
minimum = 100000
puts if debug == 1

# (1) 増やす、減らすのみのパターン
choices = bamboos.combination(3).to_a
choices.each do |choice|
  mp = count_mp(needs, choice, 0)
  minimum = mp if mp < minimum
end
puts minimum if debug == 1

# (2) 1度だけ結合を使うパターン
choices = bamboos.combination(4).to_a
choices.each do |choice|
  patterns = choice.combination(2).to_a # 合成用素材
  patterns.each do |pattern|
    new_choice = choice - pattern
    new_choice << pattern.inject(:+)
    new_choice.sort!
    mp = count_mp(needs, new_choice, 1)
    minimum = mp if mp < minimum
  end
end
puts minimum if debug == 1

# (3) 2度結合を使い、3本組のを1つ作るパターン
patterns = bamboos.combination(3).to_a
patterns.each do |pattern|
  restall = bamboos - patterns # 選ばなかった残りがrestallに入る
  rests = restall.combination(2).to_a
  rests.each do |rest|
    choice = rest.dup
    choice << pattern.inject(:+) # これで3つ目がchoiceに入る
    choice.sort!
    mp = count_mp(needs, choice, 2)
    minimum = mp if mp < minimum
  end
end
puts minimum if debug == 1

# (4) 2度結合を使い、2本組のを2つ作るパターン
bamboos.each do |solo|
  # choice = []
  # choice << solo # 合成しない1つ目
  rest = bamboos - [solo] # 残りは4つ
  as = rest.combination(2).to_a
  as.each do |combi_a|
    restall2 = rest - combi_a
    rests2 = restall2.combination(2).to_a
    rests2.each do |combi_b|
      choice = [solo, combi_a.inject(:+), combi_b.inject(:+)]
      choice.sort!
      mp = count_mp(needs, choice, 2)
      minimum = mp if mp < minimum
    end
  end
end
puts minimum if debug == 1

puts minimum