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 sys
def de(x, y):
	if (x % y == 0):
		 return x // y
	return x // y + 1

def euc(a, b):
	if (b == 0):
		return 1, 0
	x, y = euc(b, a % b)
	return -y, -x - y * (a // b)



def solve( x, y, p, q):
	if (q == p):
		if (x == y):
			 return 0
		return -1
	if (p == 0):
		if (x == 0):
			 return 0
		return -1
	
		
	a0, b0 = euc(p, q)
	g = a0 * p - q * b0
	c = x * q - p * y
	if (c % g):
		return -1
	a0 = a0 * (c // g)
	b0 = b0 * (c // g)
	t1 = a0 // q
	
	t = max(de(b0 - a0, q - p), de(-b0, p))
	a = a0 + q * t
	b = b0 + p * t

	return a


#sys.stdin = open('input.txt', 'r')


t = int(input())
for it in range(t):
	x, y, p, q = map(int, input().split())
	print(solve(x, y, p, q))
