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 math as mt 
import sys,string,bisect
input=sys.stdin.readline
import random
from collections import deque,defaultdict
L=lambda : list(map(int,input().split()))
Ls=lambda : list(input().split())
M=lambda : map(int,input().split())
I=lambda :int(input())
d=defaultdict(int)
def extended_gcd(a,b):
    if(a==0):
        return(0,1)
    x,y=extended_gcd(b%a,a)
    return (y-(b//a)*x,x)
def check(x,y,p,q):
    if(p>=x and (y-x)+p<=q):
        return True
    return False

t=I()
for _ in range(t):
    x,y,p,q=M()
    if(p==0 and x!=0):
        print(-1)
    
    else:
        g=mt.gcd(x,y)
        if(x//g==p and y//g==q):
            print(0)
        elif(p==q):
            print(-1)
        else:
            l=0
            r=1
            while(not(check(x,y,p*r,q*r))):
                l=r
                r*=10
            #print("p",l,r)
            d=0
            while(l<=r):
                m=(l+r)//2
                if(check(x,y,p*m,q*m)):
                    d=m
                    r=m-1
                else:
                    l=m+1
            if((q*d-y)<0):
                print(-1)
            else:
                print(q*d-y)
    

