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 Kotlin (judge verdict: Accepted)

fun main(args: Array<String>) {
    val num = readLine()!!.split(" ").map { it.toInt() }
    val dice = DiceB(num[0], num[1], num[2], num[3], num[4], num[5])
    val count = readLine()!!.toInt()
    for (i in (0 until count)) {
        val tempList = readLine()!!.split(" ").map { it.toInt() }
        println(dice.getRight(top = tempList[0], front = tempList[1]))
    }
}

class DiceB(
    private val one: Int,
    private val two: Int,
    private val three: Int,
    private val four: Int,
    private val five: Int,
    private val six: Int
) {

    fun getRight(top: Int, front: Int): Int {
        return when (top) {
            one -> when (front) {
                two -> three
                three -> five
                four -> two
                else -> four
            }
            two -> when (front) {
                one -> four
                three -> one
                four -> six
                else -> three
            }
            three -> when (front) {
                one -> two
                two -> six
                five -> one
                else -> five
            }
            four -> when (front) {
                one -> five
                two -> one
                five -> six
                else -> two
            }
            five -> when (front) {
                one -> three
                three -> six
                four -> one
                else -> four
            }
            else -> when (front) {
                two -> four
                three -> two
                four -> five
                else -> three
            }
        }
    }
}