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 Main {
    public static void main (String[] args) throws java.lang.Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int t = Integer.parseInt(st.nextToken());
        for (int i=0 ; i<t ; i++) {    
            st = new StringTokenizer(br.readLine());
            long x = Long.parseLong(st.nextToken()),
                y = Long.parseLong(st.nextToken()),
                p = Long.parseLong(st.nextToken()),
                q = Long.parseLong(st.nextToken());
            if (p==q && x!=y) {
                System.out.println(-1);
                continue;
            } else if (p==0 && x!= 0) {
                System.out.println(-1);
                continue;
            } else if (p==0 && x==0) {
                System.out.println(0);
                continue;
            } 

            long min = Long.MAX_VALUE;
            long l =0,
                r = 2000000000;
            while (l != r) {
                //    System.out.println(l + " " + r);
                long mid = (r+l) / 2;
                long numerator = (long)p * mid,
                    denominator = (long)q * mid;
                long a = numerator - x;
                long b = denominator - y - a;
                if (a>=0 && b>=0) {
                    r = mid;
                    min = Math.min(a+b, min);
                    continue;
                } 
                l = mid + 1;
            }
            System.out.println(min);
        }
        
    }
    
}

