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 solve():
    x, y, p, q = map(int, raw_input().split())

    if p == q and y != y:
        print "-1"
        return
    
    if p == 0:
        if x == 0:
            print 0
        else:
            print -1
        return

    if ( x%p == 0 ) and (y%q == 0) and (x/p == y/q):
        print 0
        return

    n = bsearch(x,y,p,q)

    ans = -1;
    for nn in range( max(n-3, 1), n+3 ):
        if (p * nn - x >= 0 and q * nn - y >= 0 and ((p * nn - x) <= (q * nn - y) )):
            if ans == -1:
                ans = q * nn - y
            else:
              ans = min(ans, q * nn - y)
#    print p*n, '/', q*n, '-', p, '/', q
    print ans


def bsearch(x,y,p,q):
    lo = 1
    hi = int(1e20)
    guess = -1

    while lo <= hi:
        mid = (lo + hi) / 2

        if ( (q * mid <= y) or (p * mid <= x) or ((p * mid - x) > (q * mid - y) )): 
          lo = mid + 1;
        else:
          hi = mid - 1;
          guess = mid;
    return guess;
    








t = int(raw_input())
for i in range(t):
    solve()
