We have three groups of cogs. Cogs in the same group rotate at the same radian rate as they all have the same center and be bound together. Now we connect the three groups with two chains. One from the first to the second, one from the second to the third. The two chains can not be connected to the same cog in the second group.

Now if we regard the rotation rate of the first cog group to be 1, depends on different configurations of chains, we can get different rate of the last cog group, which is a fractional number . Given the size of each cogs of the three groups, you are asked to decide if a specified fraction is able to be achieved.

We know that if the size of the cog in the first group is , it is connected to cog of size in the second group and the size of the cog in the last group is and it is connected to a cog of size . Then we have

So first we use two loop to decide and find if there exists and respectively in that two groups to make such a fraction. Also, we have to use gcd function to find the greatest common divisor and simplify the fraction. To fastly check the the existens, we first sort sizes of cogs in each group and once we get that fraction in its simplest representation, we multiply the numerator and denominator by 1, 2, 3 and so on until one of them exceed the range of cogs. Each time, we use binary search to check if we can find such and .

An other method might be that we iterate every possible combinations of sizes from the first and last cog groups and saved the in a HashSet or put them into an array and sort them. Now we can just use the set or binary search to check existence.

import java.util.*;

public class gWheels {
    private static long gcd(long a, long b) {
        if (b == 0) return a;
        else return gcd(b, a % b);
    private static void solve(long p, long q, long[] pwheel, long[] ewheel, long[] twheel) {
        long pqg = gcd(p, q);
        p /= pqg;
        q /= pqg;
        long pmax = pwheel[pwheel.length - 1];
        long tmax = twheel[twheel.length - 1];
        for (int i = 0; i < ewheel.length; i++) {
            for (int j = 0; j < ewheel.length; j++) {
                if (i == j) continue;
                long a = ewheel[i] * p;
                long b = ewheel[j] * q;
                long abg = gcd(a, b);
                a /= abg;
                b /= abg;
                for (int k = 1; k * a <= pmax && k * b <= tmax; k++) {
                    if (Arrays.binarySearch(pwheel, k * a) >= 0 && Arrays.binarySearch(twheel, k * b) >= 0) {
    public static void main(String[] args) {
        Scanner in = new Scanner(;
        int TEST = in.nextInt();
        for (int t = 0; t < TEST; t++) {
            int P = in.nextInt(), E = in.nextInt(), T = in.nextInt();
            long[] pwheel = new long[P];
            long[] ewheel = new long[E];
            long[] twheel = new long[T];
            for (int i = 0; i < P; i++) {
                pwheel[i] = in.nextLong();
            for (int i = 0; i < E; i++) {
                ewheel[i] = in.nextLong();
            for (int i = 0; i < T; i++) {
                twheel[i] = in.nextLong();
            System.out.printf("Case #%d:\n", t + 1);
            int M = in.nextInt();
            for (int i = 0; i < M; i++) {
                long p = in.nextLong(), q = in.nextLong();
                solve(p, q, pwheel, ewheel, twheel);