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, x, y) = gcd(b % a, a)
	return (g, y - (b // a) * x, x) 


t = int(input())

for i in range(t):
	(x, y, p, q) = list(map(int, input().split()))
	val = p * y - q * x
	(g, a, b) = gcd(q, -p)

	if (val % g != 0):
		print(-1)
		continue

	a *= val // g
	b *= val // g

	da = abs(p // g)
	db = abs(q // g)

	if (a < 0):
		if (da == 0):
			print(-1)
			continue

		mul = (abs(a) + da - 1) // da
		a += mul * da
		b += mul * db


	if (b < 0):
		if (db == 0):
			print(-1)
			continue

		mul = (abs(b) + db - 1) // db;
		a += mul * da
		b += mul * db

	if (a > b):
		if (da == db):
			print(-1)
			continue

		diff = a - b
		step = db - da
		mul = (diff + step - 1) // step
		a += mul * da
		b += mul * db

	mul = b // db;
	if (da != 0):
		mul = min(mul, a // da)

	if (da != db):
		mul = min(mul, (b - a) // (db - da))

	print(b - db * mul) 
	