You are an experienced Codeforces user. Today you found out that during your activity on Codeforces you have made y submissions, out of which x have been successful. Thus, your current success rate on Codeforces is equal to x / y.

Your favorite rational number in the [0;1] range is p / q. Now you wonder: what is the smallest number of submissions you have to make if you want your success rate to be p / q?

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

Each of the next t lines contains four integers x, y, p and q (0 ≤ x ≤ y ≤ 109; 0 ≤ p ≤ q ≤ 109; y > 0; q > 0).

It is guaranteed that p / q is an irreducible fraction.

Hacks. For hacks, an additional constraint of t ≤ 5 must be met.

Output

For each test case, output a single integer equal to the smallest number of submissions you have to make if you want your success rate to be equal to your favorite rational number, or -1 if this is impossible to achieve.

Example

Input

4
3 10 1 2
7 14 3 8
20 70 2 7
5 6 1 1


Output

4
10
0
-1

Note

In the first example, you have to make 4 successful submissions. Your success rate will be equal to 7 / 14, or 1 / 2.

In the second example, you have to make 2 successful and 8 unsuccessful submissions. Your success rate will be equal to 9 / 24, or 3 / 8.

In the third example, there is no need to make any new submissions. Your success rate is already equal to 20 / 70, or 2 / 7.

In the fourth example, the only unsuccessful submission breaks your hopes of having the success rate equal to 1.
    /*
    ID: mchensd
    LANG: JAVA
    PROG: SuccessRate
     */

    /**
     *
     * @author Michael
     */

    import java.util.*;
    import java.io.*;

    public class SuccessRate {
        public boolean can(long x, long y, long p, long q, long k) {
            return (x <= p*k && y <= q*k && p*k-x <= q*k-y);
        }
        public void run(BufferedReader br, PrintWriter pw) throws IOException {
            StringTokenizer st;
            int t = Integer.parseInt(br.readLine());
            for (int i=0; i<t; ++i) {
                st = new StringTokenizer(br.readLine());
                long x = Long.parseLong(st.nextToken());
                long y = Long.parseLong(st.nextToken());
                long p = Long.parseLong(st.nextToken());
                long q = Long.parseLong(st.nextToken());

                if (p == q) {
                    if (x == y) pw.println(0);
                    else pw.println(-1);
                    continue;
                }
                
                if (p == 0) {
                    if (x == 0) pw.println(0);
                    else pw.println(-1);
                    continue;
                }

                long lb = 1;
                long ub = (long) 1e10;
                while (true) {
                    if (lb == ub) break;
                    long xi = (lb + ub) / 2;
    //                System.out.println(xi);
                    if (can(x, y, p, q, xi)) ub = xi;
                    else lb = xi + 1;
                }
                pw.println(lb*q-y);
            }

        }
        public static void main(String[] args) throws IOException {
    //        new SuccessRate().runFileIO("SuccessRate");
            new SuccessRate().runStdIO();

        }

        public void runFileIO(String filename) throws IOException {
            BufferedReader br = new BufferedReader(new FileReader(filename + ".in"));
            PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter(filename + ".out")));
            run(br, pw);
            pw.close();
        }

        public void runStdIO() throws IOException {
            BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
            PrintWriter pw = new PrintWriter(System.out);
            run(br, pw);
            pw.close();
        }
    }

