Codeforces Round #412 (rated, Div. 2, base on VK Cup 2017 Round 3) C. Success Rate(数学，二分)

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.

题目大意:目前你的胜率是x/y，同时你更喜欢你的胜率是p/q，为了满足这一个要求，要求算出最小的提交次数

解题思路:根据这个要求可以得到一组公式

x+a==k*p

y+b==k*q

其中a为提交成功率，b为提交量，然后通过二分获得答案、

#include<iostream>
#include<cstdio>
#include<stdio.h>
#include<cstring>
#include<cstdio>
#include<climits>
#include<cmath>
#include<vector>
#include <bitset>
#include<algorithm>
#include <queue>
#include<map>
using namespace std;

/*start:17/5/19 14:54*/
/*end:17/5/19   14:37  */
long long int p,q,x,y,l,r,k,a,b;
int T;

int main()
{
cin>>T;
while(T--)
{
cin>>x>>y>>p>>q;
l=0;
r=1e10;
while(l<r)
{
k=(l+r)/2;
a=k*p-x;
b=k*q-y;
if(a>=0&&b>=0&&a<=b)
{
//cout<<1111<<endl;
r=k;
}
else
{
l=k+1;
}
}
//cout<<r<<endl;
if(l==1e10)
cout<<-1<<endl;
else
cout<<(r*q-y)<<endl;
}
}