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.

        
    
def eulid(p,q):
    #p<q
    A=[1,0]
    B=[0,1]
    while(p!=1 and q!=1 ):
        if q<0:
            q=-q
            B[0]=-B[0]
            B[1]=-B[1]
        if p<0:
            p=-p
            A[0]=-A[0]
            A[1]=-A[1]
        
        
        if q<p:
            p,q=q,p
            A,B=B,A
        #always assume p<q
        T=q//p
        q=q-T*p
        B[0]=B[0]-T*A[0]
        B[1]=B[1]-T*A[1]
        
        
        if q<0:
            q=-q
            B[0]=-B[0]
            B[1]=-B[1]
        if p<0:
            p=-p
            A[0]=-A[0]
            A[1]=-A[1]
    if p==1:
        return A
    elif q==1:
        return B
        

def find(x,y,p,q):
    if p==q:
        if x==y:
            return 0
        else:
            return -1
    elif p==0:
        if x==0:
            return 0
        else:
            return -1
    elif p*y==x*q:
        return 0
    else:
        A=x*q-p*y
        a,b=eulid(p,q)
        
        k0,t0=A*a,-b*A
        temp=t0%p
        r=(temp-t0)//p
        k1=k0+r*q
        t1=t0+r*p
        if k1>=t1:
            return k1
        else:
            if (t1-k1)%(q-p)==0:
                r=(t1-k1)//(q-p)
            else:
                r=(t1-k1)//(q-p)+1
            k2=k1+r*q
            t2=t1+r*p
            return k2
            
            

        

    
    
T=int(input())
for _ in range(T):
    x,y,p,q=list(map(int,input().strip().split(' ')))
    print(find(x,y,p,q))
    
        
        
        
         
    

 
    