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.
t=input()
from fractions import *
from bisect import *
for _ in range(0,t):
    x, y, p, q = map(int, raw_input().split())


    if 1==1:

        if x*q==p*y:
            print 0

        else:
            f=[]
            u=10**14
            l=0
            #x,y,p,q=p,q,x,y
         #   print x,y,p,q
            for i in range(0,300):
                t=(u+l)/2
                ux=p*t
                uy=q*t
                sx2=p*t-x
                ux2=q*t-sx2-y

                if sx2<0 or ux2<0:
                    l=t+1
                else:
                    u=t-1
                #if sx2*1.0==p*(ux2+sx2)*1.0/q:
                    f.append(sx2+ux2)
               # print l,u,t,sx2,ux2,ux,uy
                #print i,val,y,t,l,u,t
           # print f
            if len(f)==0:
                print -1
            else:
                print min(f)