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.
# (x + a) / (y + a + b) = p / q
# minimize a+b
# (x + a) * q = (y + a + b) * p
# (3 + a) * 2 = (10 + a + b) * 1
# xq + aq = yp +ap +bp
# xq - yp = ap + bp - aq
# xq - yp = (a+b)p -aq
# xq - yp = (p-q)a +bp
# 3*2 - 10*1 = (a+b)1 -2a

#(p-q)a+bp=left
#b=(left-(p-q)a)/p
# left+aq=(a+b)p

import math

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

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

	lo = 0
	hi = 2**65

	while hi-lo > 1:
		#print(lo,hi)
		mid = (lo+hi)//2
		k = mid

		if x + k*q - y >= k*p and x <= k*p and k*q >= y:
			hi = mid
		else:
			lo = mid

	print(q*hi-y)