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):
    while a != 0 and b != 0:
        if a > b:
            a = a % b
        else:
            b = b % a
    return a + b


def check(x,y,p,q,m):
    num = p * m
    den = q * m
    return (num - x >= 0) and ((den - y - (num - x)) >= 0)

def count(x,y,p,q,m):
    num = p * m
    den = q * m
    return num - x + (den - y - (num - x))
    

def search(x,y,p,q):
    l,r = -1, 10 ** 9 + 1
    while r - l > 1:
        m = (l + r) // 2
        if check(x,y,p,q,m):
            r = m
        else:
            l = m
    if r == 10 ** 9 + 1:
        print(-1)
    else:
        print(count(x,y,p,q,r))
    
    


n = int(input())
for i in range(n):
    x,y,p,q = map(int, input().split())
    search(x,y,p,q)