Problem:
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Dice II

Construct a dice from a given sequence of integers in the same way as Dice I.

You are given integers on the top face and the front face after the dice was rolled in the same way as Dice I. Write a program to print an integer on the right side face.

Input

In the first line, six integers assigned to faces are given in ascending order of their corresponding labels.
In the second line, the number of questions $q$ is given.

In the following $q$ lines, $q$ questions are given. Each question consists of two integers on the top face and the front face respectively.

Output

For each question, print the integer on the right side face.

Constraints

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

The integers are all different

$1 \leq q \leq 24$

Sample Input

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

Sample Output

3
5
6

Solution in Ruby (judge verdict: Time Limit Exceeded)

dice = gets.split.map(&:to_i)
n = gets(chomp: true).to_i
order = [].tap{|a|n.times{a << gets.split.map(&:to_i)}}

def vertical_roll(new_dice, upper)
  while new_dice[0] != upper
    new_dice[0], new_dice[1], new_dice[4], new_dice[5] = new_dice[1], new_dice[5], new_dice[0], new_dice[4]
  end
  new_dice
end

def horizontal_roll(new_dice, front)
  while new_dice[1] != front
    new_dice[1], new_dice[2], new_dice[4], new_dice[3] = new_dice[2], new_dice[4], new_dice[3], new_dice[1]
  end
  new_dice
end

order.each do |odr|
  new_dice = dice
  idx1 = dice.index(odr[0])
  idx2 = dice.index(odr[1])
  case idx1
  when 0,1,4,5
    case idx2
    when 0,1,4,5
      vertical_roll(new_dice, odr[0])
    else
      horizontal_roll(vertical_roll(new_dice, odr[0]), odr[1])
    end
  else
    vertical_roll(horizontal_roll(new_dice, odr[1]), odr[0])
  end
  puts new_dice[2]
end