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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class R412C
{
	static long x;
	static long y;
	static long p;
	static long q;
	public static void main(String[] args) throws IOException
	{
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		
		int tc = sc.nextInt();
		while(tc-- > 0)
		{
			x = sc.nextLong();
			y = sc.nextLong();
			p = sc.nextLong();
			q = sc.nextLong();
			
			if(p == 0)
			{
				if(x == 0)
					pw.println(0);
				else
					pw.println(-1);
				continue;
			}
			
			long mid = search(1, (long)1e9+10);
			long dom = mid * q;
			long nom = mid * p;
			if(mid == (long)1e9+10)
				pw.println(-1);
			else
				pw.println(dom - y);
		}
		
		pw.flush();
		pw.close();
	}
	
	private static long search(long st, long en)
	{
		if(st == en)
			return st;
		long mid = st + (en-st)/2;
		long dom = mid * q;
		long nom = mid * p;
		if(dom < y || nom < x)
			return search(mid+1, en);
		if(dom - y >= nom - x)
			return search(st, mid);
		else
			return search(mid+1, en);
	}

	static class Scanner 
	{
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}

		public String next() throws IOException 
		{
			while (st == null || !st.hasMoreTokens()) 
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public int nextInt() throws IOException {return Integer.parseInt(next());}
		
		public long nextLong() throws IOException {return Long.parseLong(next());}

		public String nextLine() throws IOException {return br.readLine();}
		
		public double nextDouble() throws IOException
		{
			String x = next();
			StringBuilder sb = new StringBuilder("0");
			double res = 0, f = 1;
			boolean dec = false, neg = false;
			int start = 0;
			if(x.charAt(0) == '-')
			{
				neg = true;
				start++;
			}
			for(int i = start; i < x.length(); i++)
				if(x.charAt(i) == '.')
				{
					res = Long.parseLong(sb.toString());
					sb = new StringBuilder("0");
					dec = true;
				}
				else
				{
					sb.append(x.charAt(i));
					if(dec)
						f *= 10;
				}
			res += Long.parseLong(sb.toString()) / f;
			return res * (neg?-1:1);
		}
		
		public boolean ready() throws IOException {return br.ready();}
	}
}