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 xgcd(a,b):
  if b==0: return [a,1,0]
  g,x1,y1=xgcd(b,a%b)
  x=y1
  y=x1-(a/b)*y1
  return [g,x,y]
def cel(a,b):
  if b<0: 
    a*=-1
    b*=-1;
  if a>0: return (a+b-1)/b;
  ans=(-a)/b;
  return -ans;

def solve(a,b,c):
  #a*x+b*y=c, x>=0, y<=0, min x+y
    if a==0: 
        print ("-1")
        return
    if b==0:
        if c%a==0: print c/a
        else: print "-1"
        return
    g,x,y=xgcd(a,b)
    x*=c
    y*=c
    k=max(cel(-x,b),cel(y,a))
    ans=x-y+k*(b+a);
    print ans
t=input()
for tt in range(t):
    x,y,p,q=map(long,raw_input().split())
    if p==0:
        if x==0: print "0"
        else: print "-1"
    elif p==q:
        if x==y: print "0";
        else: print "-1";
    elif p*y-x*q>0: solve(q-p,p,p*y-x*q);
    else: solve(p,q-p,x*q-p*y);
