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 mcd(a, b):
  if(a == 0):
    return b
  return mcd(b % a, a)

def mcde(a, b):
  r = [0,0,0]
  if(b == 0):
    r[0] = a
    r[1] = 1
    r[2] = 0
  else:
    t = mcde(b,a%b);
    r[0] = t[0]
    r[1] = t[2];
    r[2] = t[1] - a/b*t[2]
  return r


t = int(raw_input())

for caso in range(t):
  x,y,p,q = map(int, raw_input().split())
  if(p == 0):
    if(x == 0):
      print 0
    else:
      print -1
    continue
    
  if(p == q):
    if(x == y):
      print 0
    else:
      print -1
    continue
  
  a = x * q - p * y
  r = mcde(p,q)

  l = a * r[1]
  k = - a * r[2]
  
  if(k < 0):
    su = (-k + p - 1) / p
  else:
    su = k / p
    su = -su
  #inf = (k - l) / (q - p);
  if(k > l):
    inf = (k - l + q - p - 1) / (q - p)
  else:
    inf = (l - k) / (q - p)
    inf = -inf
  n = max(inf,su)
  cnd = l + n * q

  print cnd
