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 math

def haha():
    n = int(raw_input())
    for i in xrange(n):
        x = raw_input()
        x = x.split(" ")
        a = int(x[0])
        b = int(x[1])
        c = int(x[2])
        d = int(x[3])
        e = b - a
        f = d - c
        k = float(c)/float(d)
        if k == float(1):
            if a == b:
                print 0
            else: 
                print -1
        elif k == 0.0:
            if a == 0:
                print 0
            else:
                print -1
        elif k > 1.0 or k < 0.0:
            print -1
        else:
            t = max(math.ceil(float(e)/float(f)),math.ceil(float(a)/float(c)))
            t = int(t)
            print (c+f)*t - a - e
haha()