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

public class A {
    FastScanner in;
    PrintWriter out;

    public void solve() throws IOException {
        int tests = in.nextInt();
        for (int i = 0; i < tests; i++) {
            solveTest();
        }
    }

    private void solveTest() {
        long x = in.nextInt();
        long y = in.nextInt();
        long p = in.nextInt();
        long q = in.nextInt();
        if (p == 0) {
            if (x == 0) {
                out.println(0);
            } else {
                out.println(-1);
            }
        } else {
            if (p == q) {
                if (x == y) {
                    out.println(0);
                } else {
                    out.println(-1);
                }
            } else {
                long a = Math.max((y + q - 1) / q, (x + p - 1) / p);
                long nx = a * p;
                long ny = a * q;
                if (nx - x <= ny - y) {
                    out.println(ny - y);
                } else {
                    long b = nx - x - (ny - y);
                    b = (b + q - p - 1) / (q - p);
                    out.println(ny - y + b * q);
                }
            }
        }
    }

    public void run() {
        try {
            in = new FastScanner(System.in);
            out = new PrintWriter(System.out);
            Locale.setDefault(Locale.US);

            solve();

            out.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(InputStream f) {
            br = new BufferedReader(new InputStreamReader(f));
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
    }

    public static void main(String[] arg) {
        new A().run();
    }
}