Problem:
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Dice III

Write a program which reads the two dices constructed in the same way as Dice I, and determines whether these two dices are identical. You can roll a dice in the same way as Dice I, and if all integers observed from the six directions are the same as that of another dice, these dices can be considered as identical.

Input

In the first line, six integers assigned to faces of a dice are given in ascending order of their corresponding labels.

In the second line, six integers assigned to faces of another dice are given in ascending order of their corresponding labels.

Output

Print "Yes" if two dices are identical, otherwise "No" in a line.

Constraints

$0 \leq $ the integer assigned to a face $ \leq 100$

Sample Input 1

1 2 3 4 5 6
6 2 4 3 5 1

Sample Output 1

Yes

Sample Input 2

1 2 3 4 5 6
6 5 4 3 2 1

Sample Output 2

No

Solution in Ruby (judge verdict: Accepted)

dice1 = gets.split.map(&:to_i)
dice2 = gets.split.map(&:to_i)
throw = {E: [3, 2, 6, 1, 5, 4], N: [5, 1, 3, 4, 6, 2],
         S: [2, 6, 3, 4, 1, 5], W: [4, 2, 1, 6, 5, 3]}
th = {}
throw.each_key {|k| th[k] = throw[k].map(&:pred)}

front = dice1[1]
unless dice2.find_index(front)
  puts "No"
  exit
end

rotate = ->(dir, dice) {
  next_dice = Array.new(6, 0)
  dice.each_with_index do |d, i|
    next_dice[th[dir][i]] = d
  end
  next_dice
}

stack = []
dice = dice2
next_dice = loop do
  [:E, :N, :S, :W].each {|d| stack << [dice, d]}
  dice, dir = stack.shift
  next_dice = rotate.(dir, dice)
  if next_dice[1] == front
    break next_dice
  else
    dice = next_dice
  end
end

result = "No"
4.times do
  if next_dice == dice1
    result = "Yes"
    break
  end
  next_dice = rotate.(:E, next_dice)
end
puts result