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.
t = int(input())

for tc in range(t):

    x, y, p, q = map(int, input().split())

    largenum = 10**18

    c12 = ((q*largenum - y) >= (p*largenum - x))
    c22 = (p*largenum >= x)
    c32 = (q*largenum >= y)

    if not (c12 and c22 and c32):
        print(-1)
        continue

    left, right = 0, 10**18

    while left + 1 < right:
        mid = (left + right) // 2

        c1 = ((q*mid - y) >= (p*mid - x))
        c2 = (p*mid >= x)
        c3 = (q*mid >= y)

        if c1 and c2 and c3:
            right = mid
        else:
            left = mid

    ans = right

    print(int(ans * q - y))


