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.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
import java.io.IOException;


public class JavaApplication6 {
static long ans;
static long c;

public static long nextmq(long y, long q)
{
    return ((y-1)/q)*q+q;
}
//public static void solve(long x, long y, long p, long q)
//{
//    
//}

    public static void main(String[] args) throws IOException{
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter pw = new PrintWriter(System.out);
    StringTokenizer st = new StringTokenizer(br.readLine());

    int t = Integer.parseInt(st.nextToken());
    for(int i =0;i<t;i++)
    {    
    StringTokenizer st2 = new StringTokenizer(br.readLine());
    long x = Integer.parseInt(st2.nextToken());
    long y = Integer.parseInt(st2.nextToken());
    long p = Integer.parseInt(st2.nextToken()); 
    long q = Integer.parseInt(st2.nextToken());
    if(p==0)
    {
        if(x==0) 
        {ans = 0;
        pw.println(ans);
        continue;}
        else {ans = -1;    
        pw.println(ans);
        continue;}
    }
    if(p==q)
    {
        if(x==y) {ans = 0;
                pw.println(ans);
        continue;}
        else 
        {ans = -1;
        pw.println(ans);
        continue;}
    }
    if (p*y==x*q){ans =0; pw.println(ans);continue;}
    if(x*q<y*p)
    {
        if(q-p==1) c = y*p-x*q;
        else
        {c = (y*p-x*q-1)/(q-p)+1;}
        long y2 = nextmq(y+c,q);
        //long x2= (y2/q)*p;
        ans = (y2-y);
        pw.println(ans);
        continue;   
    }
    if(x*q>y*p)
    {
        long y2 = 0;
        if (p==1) y2= x*q;
        else
        {y2=(x*q-1)/p+1;}
        long y3=nextmq(y2,q);
        //long x3 = (y3/q)*p;
        ans = (y3-y);
        pw.println(ans);
    }
   
    
    }
    pw.close();
    }
    
}