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.lang.*;
import java.util.*;

public class A773 {
    public static void main(String[] args) throws IOException{
        StringBuffer ans = new StringBuffer();
        StringTokenizer st;
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        st = new StringTokenizer(f.readLine());
        int t = Integer.parseInt(st.nextToken());
        for(; t > 0; t--){
            st = new StringTokenizer(f.readLine());
            long x = Integer.parseInt(st.nextToken());
            long y = Integer.parseInt(st.nextToken());
            long p = Integer.parseInt(st.nextToken());
            long q = Integer.parseInt(st.nextToken());
            long r = binarySearch(x,y,p,q);
            if(!works(x,y,p*r,q*r)) ans.append(-1).append("\n");
            else
                ans.append(r*q - y).append("\n");
        }
        f.close();



        System.out.println(ans);
    }
    public static boolean works(long x, long y, long p, long q){
        return p >= x && q >= y && q-y >= p-x;
    }
    public static long binarySearch(long x, long y, long p, long q){
        long lo = 0;
        long hi = (long) Math.pow(10,9);
        while (lo + 1 < hi) {
            long mid = (lo + hi)/2;
            //System.out.println(mid + " " + works(x,y,p*mid,q*mid));
            if (!works(x,y,p*mid,q*mid)) {
                lo = mid;
            } else {
                hi = mid;
            }
        }
        return hi;
    }
}