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
import sys

input_ = sys.stdin.readline


def solve3(x, y, p, q):
    if p == q:
        if x == y:
            return 0
        return -1

    if p == 0:
        if x == 0:
            return 0
        return -1

    diff = (q - y % q) % q
    m = (y + diff) / q

    n1 = (x - p * m) / p
    n2 = (m * p - x - diff) / (q - p)
    n = math.ceil(max(n1, n2))

    return diff + n * q


def main():
    t = int(input_())
    solutions = []

    for _ in range(t):
        x, y, p, q = map(int, input_().split())
        solutions.append(solve3(x, y, p, q))

    return solutions

if __name__ == '__main__':
    solutions = main()

    for s in solutions:
        print(s)
