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 sys

def exgcd(a, b):
    d = a; x = 1; y = 0
    if b != 0:
        d, y, x = exgcd(b, a % b)
        y -= (a / b) * x
    return (d, x, y)

def solve(a, b, c):
    if a == 0 and b == 0: 
        if c != 0: return -1
        else: return 0
    if a == 0: 
        if c % b == 0 and c / b <= 0: return -c / b
        else: return -1
    if b == 0:
        if c % a == 0 and c / a >= 0: return c / a
        else: return -1

    g, x, y = exgcd(a, b)
    if c % g != 0: return -1

    a /= g; b /= g; c /= g
    x *= c; y *= c

    if x > 0:
        k = x / b
        x -= k * b
        y += k * a
    elif x < 0:
        k = (-x + b - 1) / b
        x += k * b
        y -= k * a
    if y > 0:
        k = (y + a - 1) / a
        x += k * b
        y -= k * a
    return x - y

fin = sys.stdin
#fin = open(r'C:\Users\TaoSama\Desktop\in.txt')
fout = sys.stdout

t = int(fin.readline())
for _ in range(t):
    x, y, p, q = map(int, fin.readline().split())
    fout.write("{}\n".format(solve(q - p, p, p * y - q * x)))