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 sys
from fractions import gcd as gcdlib

def main():
    t=int(sys.stdin.readline().rstrip())
    for _ in range(t):
        x,y,p,q=list(map(int,sys.stdin.readline().split()))
        
        res,m,k=find_best_solution(p-q,p,-(y*p-x*q))
        if res==False: result=-1
        elif m<0 or k<0: result=-1
        else: result=m+k
        
        sys.stdout.write(str(result)+'\n')
    
def gcdex(a, b):
    if b == 0:
        return a, 1, 0
    else:
        d, x, y = gcdex(b, a % b)
        return d, y, x - y * (a // b)
    
def gcd(a,b):
    if a==0:
        return b,0,1
    d,x,y=gcd(b%a, a)
    return d, y-(b//a)*x, x
    
def find_any_solution(a,b,c):
    g,x0,y0=gcd(abs(a), abs(b))
    if c%g!=0: return False,None,None
    x0*=c//g
    y0*=c//g
    if a<0: x0*=-1
    if b<0: y0*=-1
    return True,x0,y0
    
def find_best_solution(a,b,c):
    res,x0,y0=find_any_solution(a,b,c)
    if res==False: return res,x0,y0
    
    g=abs(gcdlib(a,b))
    
    if a==0:
        k=-x0//(b//g)
    elif b==0:
        k=-y0//(abs(a)//g)
    else:
        k=-min(x0//(b//g), y0//(abs(a)//g))
    
    x=x0+k*(b//g)
    y=y0-k*(a//g)
    
    return True,x,y
    
main()

