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 mp():  return map(int,input().split())
def lt():  return list(map(int,input().split()))
def pt(x):  print(x)
def ip():  return input()
def it():  return int(input())
def sl(x):  return [t for t in x]
def spl(x): return x.split()
def aj(liste, item): liste.append(item)
def bin(x):  return "{0:b}".format(x)
def listring(l): return ' '.join([str(x) for x in l])
def ptlist(l): print(' '.join([str(x) for x in l]))

t = it()
for _ in range(t):
    x,y,p,q = mp()
    if p == q:
        if x == y:
            pt(0)
        else:
            pt(-1)
    elif p == 0:
        if x == 0:
            pt(0)
        else:
            pt(-1)
    else:
        
        r = (y+q-1)//q
        xa,ya = r*p,r*q
        if xa < x:
            m = (x-xa+p-1)//p
        else:
            m = 0
        if xa - x > ya - y:
            n = (xa-x-(ya-y)+q-p-1)//(q-p)
        else:
            n = 0
        pt(ya+q*max(m,n)-y)
