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 gcd(a,b):
	if a==0:
		return (b, 0, 1)
	(g, x1, y1) = gcd(b%a, a) 
	x = y1-(b//a)*x1;
	y = x1;
	return (g,x,y)

 
def F(x,y,p,q):
	(dddd, A, B) = gcd(p,q)
	if(A < 0):
		t = -A//q+1
		A+=q*t;
		B-=p*t;
		while(A-q >= 0):
			A-=q;
			B+=p;
	T = x*q-p*y;
	A*=T;
	B*=T;
	B1 = -B;
	t = min(B1//p,A//q);
	return A-t*q;
 

T = int(raw_input())
for test in range(T):
	x, y, p, q = map(int, raw_input().split(' '))
	
	if p==0:
		print (0 if x==0 else -1)
		continue
	
	A = 0
	if x*q < y*p:
		if p==q:
			print -1
			continue
		A = (y*p-x*q+(q-p-1))//(q-p);
		x+=A;
		y+=A;
	print A+F(x,y,p,q)
