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

public class C {
    private final Scanner in;
    private final PrintWriter out;

    public static void main(String[] args) throws IOException {
        InputStream is = args.length > 0 ? new FileInputStream(args[0]) : System.in;
        OutputStream os = args.length > 1 ? new FileOutputStream(args[1]) : System.out;
        new C(is, os);
    }

    private C(InputStream is, OutputStream os) throws IOException {
        in = new Scanner(new BufferedReader(new InputStreamReader(is)));
        out = new PrintWriter(new OutputStreamWriter(os));
        solve();
        out.flush();
    }

    private void solve() {
        int t = in.nextInt();

        for (int i = 0; i < t; i++) {
            int x = in.nextInt();
            int y = in.nextInt();
            int p = in.nextInt();
            int q = in.nextInt();
            out.println(minTries(x, y, p, q));
        }
    }

    private long minTries(int x, int y, int p, int q) {
        if (p == 0) {
            if (x == 0) {
                return 0;
            }

            return -1;
        }

        if (p == q) {
            if (x == y) {
                return 0;
            }

            return -1;
        }

        int n0 = x / p + (x % p == 0 ? 0 : 1);
        int n1 = (y - x) / (q - p) + ((y - x) % (q - p) == 0 ? 0 : 1);
        long n = Math.max(n0, n1);

        return n * q - y;
    }
}
