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.*;
import java.math.*;

public final class part_2
{
    static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	static FastScanner sc=new FastScanner(br);
    static PrintWriter out=new PrintWriter(System.out);
	static Random rnd=new Random();
		
	static boolean check(long mid,long now,long x,long y,long p,long q)
	{
		long curr = (y + now) + ((mid-1)*q);
		
		long curr1 = (curr / q);
		
		long val1 = (p*curr1);
		
		long res = val1 - x;
		
		long ops = now + ((mid-1)*q);
		
		return (res>=0 && res<=ops);
	}
	
	static long gcd(long a,long b)
	{
		if(b==0)	
		{
			return a;
		}
		
		else
		{
			return gcd(b,a%b);
		}
	}
	
    public static void main(String args[]) throws Exception
    {
		int t=sc.nextInt();
		
		while(t>0)
		{
			long x=sc.nextLong(),y=sc.nextLong(),p=sc.nextLong(),q=sc.nextLong();
			
			if(x/gcd(x,y)==p && y/gcd(x,y)==q)
			{
				out.println(0);
			}
			
			else
			{
			
				long now = (q-(y%q))%q;

				long low=1,high=(long)(1e9)+5;

				while(low<high)
				{
					long mid=(low+high)>>1;

					if(check(mid,now,x,y,p,q))
					{
						high=mid;
					}

					else
					{
						low=mid+1;
					}
				}

				if(check(low,now,x,y,p,q))
				{
					long res = now + ((low-1)*q);out.println(res);
				}
				else
				{
					out.println(-1);
				}
			}
			
			t--;
		}
		
		out.close();
    }
}
class FastScanner
{
    BufferedReader in;
    StringTokenizer st;

    public FastScanner(BufferedReader in) {
        this.in = in;
    }
	
    public String nextToken() throws Exception {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }
	
	public String next() throws Exception {
		return nextToken().toString();
	}
	
    public int nextInt() throws Exception {
        return Integer.parseInt(nextToken());
    }

    public long nextLong() throws Exception {
        return Long.parseLong(nextToken());
    }

    public double nextDouble() throws Exception {
        return Double.parseDouble(nextToken());
    }
}