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 fractions
import sys

def dsolve(a, b, c):
    if c % fractions.gcd(a,b) != 0:
        return None
    x = a / b
    y = a % b
    if y == 0:
        return (0, c / b)
    else:
        s = dsolve(b, y, c)
        u, v = s
        return (v, u-x*v)

def solve(x, y, p, q):
    if p == 0 and x != 0:
        return (-1,-1)
    elif p == q and x != y:
        return (-1,-1)

    c = p*y-x*q
    a = q-p
    b = -p
    r1, r2 = dsolve(a, b, c)
    d = fractions.gcd(p, q)
    x_inc = b/d
    cand_a = None
    cand_b = None
    if d != 0 and b != 0:
        t1 = ((-r1)*d/b)
        a1 = t1 * (b/d) + r1 
        b1 = r2-(a/d)*t1
        if a1 >= 0 and b1 >= 0:
            cand_a = (a1, a1 + b1)
    if d != 0 and a != 0:
        t2 = ((r2*d/a))
        a2 = t2 * (b/d) + r1 
        b2 = r2-(a/d)*t2
        if a2 >= 0 and b2 >= 0:
            cand_b = (a2, a2 + b2)
    if cand_a and cand_b:
        return cand_a if cand_a[1] < cand_b[1] else cand_b
    elif cand_a:
        return cand_a
    elif cand_b:
        return cand_b
    else:
        return (-1, -1)

if __name__ == "__main__":
    T = int(sys.stdin.readline())
    for line in range(T):
        try:
            x, y = solve(*map(int, sys.stdin.readline().split()))
            print y
        except:
            print 0
