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]
	d = gcd(b % a, a)
	return [d[0], d[2] - (b // a) * d[1], d[1]]

t = int(input())
while t > 0:
	t -= 1
	x, y, p, q = map(int, input().split())
	if p == q:
		if x == y:
			print(0)
		else:
			print(-1)
		continue
	if p == 0:
		if x == 0:
			print(0)
		else:
			print(-1)
		continue
	a = p - q
	b = p
	c = q * x - p * y
	g, xa, ya = gcd(abs(a), abs(b))
	if c % g != 0:
		print(-1)
	else:
		xa *= c // g
		ya *= c // g
		if a < 0:
			xa = -xa
		if b < 0:
			ya = -ya
		if xa < 0:
			gaps = (-xa + (b // g) - 1) // (b // g)
			xa += gaps * (b // g)
			ya -= gaps * (a // g)
		if ya < 0:
			gaps = (-ya + (-a // g) - 1) // (-a // g)		
			xa += gaps * (b // g)
			ya -= gaps * (a // g)
		#print(xa, ya, a, b, c)
		if xa < 0 or ya < 0:
			print(-1)
		else:
			addon = min(xa // (b // g), ya // (-a // g))
			xa -= addon * (b // g)
			ya += addon * (a // g)
			print(xa + ya)
