Codeforces 807C Success Rate【二分+数学思维】

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本文探讨了如何通过控制新增提交的数量及成功与否来调整Codeforces平台上的个人成功率至目标比率p/q的问题。文中提出了一种有效的算法策略，通过二分查找确定最小的提交次数，以实现目标成功率。

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

Output

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：

现在你提交的正确率是x/y目标达成的正确率是p/q

现在你可以无限次提交，同时可以控制一道题的正确与否，问最少提交多少次可以达到目标正确率。

思路：

问题转换变成：

找到最小的一个值B使得：（x+A）/（y+B）==p/q.这里要保证（A<=B）;

那么此时已知：

x+A=k*p

y+B=k*q.

那么A=k*p-x，B=k*q-y..

很显然，当k越小的时候，A越小，所以这里二分k值。

然后判断B是否小于等于A即可。

同时要保证A,B>=0.

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
#define ll __int64
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        ll x,y,p,q;
        scanf("%I64d%I64d%I64d%I64d",&x,&y,&p,&q);
        ll ans=-1;
        ll l=0;
        ll r=1000000000;
        while(r-l>=0)
        {
            ll mid=(l+r)/2;
            ll A=mid*p-x;
            ll B=mid*q-y;
            if(A>=0&&B>=0&&A<=B)
            {
                ans=mid;
                r=mid-1;
            }
            else l=mid+1;
        }
        if(ans==-1)printf("-1\n");
        else 
        printf("%I64d\n",ans*q-y);
    }
}