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 R807C {
    FastScanner in;
    PrintWriter out;

    void solve() {
        int t=in.nextInt();
        while(t-->0) {
            long x=in.nextLong(), y=in.nextLong(), p=in.nextLong(), q=in.nextLong();
            if (x*q==p*y) {
                System.out.println(0);
            }
            else if ((q==1 && x!=y) || (p==0 && x>0))
                System.out.println(-1);
            else {
                long n1=x/p+((x%p!=0)?1:0);
                long n2=(y-x)/(q-p)+(((y-x)%(q-p)!=0)?1:0);
                System.out.println(Math.max(n1, n2)*q-y);
            }
        }

    }

    void run() {
        try {
            in = new FastScanner(new File("CF.in"));
            out = new PrintWriter(new File("CF.out"));

            solve();

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

    void runIO() {

        in = new FastScanner(System.in);
        out = new PrintWriter(System.out);

        solve();

        out.close();
    }

    class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner(File f) {
            try {
                br = new BufferedReader(new FileReader(f));
            } catch (FileNotFoundException e) {
                e.printStackTrace();
            }
        }

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

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

        boolean hasMoreTokens() {
            while (st == null || !st.hasMoreTokens()) {
                String s = null;
                try {
                    s = br.readLine();
                } catch (IOException e) {
                    e.printStackTrace();
                }
                if (s == null)
                    return false;
                st = new StringTokenizer(s);
            }
            return true;
        }

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

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }
    }

    public static void main(String[] args) {
        new R807C().runIO();
    }

}
