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.
def gcd(a, b):
    if b > a:
        return gcd(b, a)
    if b == 0:
        return a
    return gcd(b, a % b)

def works(a, b, c, d, x):
    return b <= d * x and a <= c * x and d * x - b >= c * x - a

def solve():
    a, b, c, d = map(int, input().rstrip().split())
    if c == d == 1 and not a / b == 1:
        print(-1)
        return
    if c == 0 and not a == 0:
        print(-1)
        return
    g = gcd(c, d)
    c //= g
    d //= g
    low = -1
    high = 1000000000000
    while low + 1 < high:
        mid = (low + high) // 2
        if works(a, b, c, d, mid):
            high = mid
        else:
            low = mid
    print(d * high - b)

if __name__ == '__main__':
    t = int(input())
    for _ in range(t):
        solve()
