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
43 10 1 27 14 3 820 70 2 75 6 1 1
Output
4100-1
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.
由题可知，要满足（x+a）/（y+b） == np/nq ，所以（x+a）=np ， （y+b）=nq。由于0 <= a <= b,所以n>=x/p,x>=(y-x)/(q-p) 向上取整求满足条件的最小n值，
就可得答案是b=nq-y；
p==q 和p==0特判下
1 #include <bits/stdc++.h>
2 using namespace std;
3 #define ll long long
4 ll x, y, q, p;
5 int main(){
6 int t;
7 cin >> t;
8 while(t--){
9 cin >> x >> y >> p >> q;
10 if(q == p){
11 printf("%d\n",(x==y)?0:-1);
12 continue;
13 }
14 if(p == 0){
15 printf("%d\n",(x==0)?0:-1);
16 continue;
17 }
18 ll GCD = __gcd(q,p);
19 q/=GCD; p/=GCD;
20 ll n = max(ceil(x*1.0/p),ceil((y-x)*1.0/(q-p)));
21 printf("%lld\n",n*q-y);
22 }
23 return 0;
24 }