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 search_k(x, y, p, q):
	lo = 0
	hi = int(1e18 + 10)

	while lo < hi:
		k = (lo+hi)//2
		test1 = p * k >= q * x - p * y
		test2 = (q-p)*k >= p * y - q * x

		if test1 and test2:
			hi = k
		else:
			lo = k+1
	k = lo
	if ((k+y)%q) > 0:
		k += (q-(k+y)%q)
	return k


t = int(input())



while t > 0:
	t -= 1
	(x, y, p, q) = input().split(' ')
	x = int(x)
	y = int(y)
	p = int(p)
	q = int(q)

	if p == 1 and q == 1:
		if x < y:
			print("-1")
		else:
			print("0")
		continue

	if p == 0:
		if x > 0:
			print("-1")
		else:
			print("0")
		continue

	print(search_k(x, y, p, q))
