CF-807C

C. Success Rate

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

题意：
我们可以将x+1，y+1或只y+1，使得x/y==p/q，问y最少要加几次，

若不能，输出-1.

由题意可得：

(x+a)/(y+b)=p/q

x+a=k*p ,  y+b=k*q

a=k*p-x ,  b=k*q-y

二分查找k，找到满足条件的最小值。

附AC代码：

 #include<bits/stdc++.h>
 using namespace std;
 
 const int INF=;
 long long int x,y,p,q;
 
 int main(){
     int t;
     cin>>t;
     while(t--){
         cin>>x>>y>>p>>q;
         long long int r=INF;
         long long int l=;
         long long int temp=-;
         while(l<=r){
             long long int mid=(l+r)/;
             long long int a=mid*p-x;
             long long int b=mid*q-y;
             if(a>=&&b>=&&a<=b){
                 temp=b;
                 r=mid-;
             }
             else{
                 l=mid+;
             }
         }
         if(temp==-)
         cout<<"-1"<<endl;
         else
         cout<<temp<<endl;
     }
     return ;
 }