Success Rate CodeForces

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.

题目链接：

CodeForces - 807C 

题意：给你4个整数，x,y,p,q，P/Q的范围是[0,1]，让你求最小的提交数量b，其中a个提交成功使 ( x + a ) / ( y + b ) == p / q

我们设一个系数n，使

p*n=x+a

q*n=y+b

那么，

a=p*n-x

b=q*n-y

根据题意，我们知道a和b满足的条件为b>=a>=0

并且观察可知a和n呈正相关，那么我们要求最小的a，可以通过二分n来得到

根据题目的数据范围，n的二分区间为0~1e9

注意下-1的情况就行了。

细节看我的AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define gg(x) getInt(&x)
using namespace std;
typedef long long ll;
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int t;
ll x,y,a,b;
int main()
{
    gbtb;
    cin>>t;
    while(t--)
    {
        cin>>x>>y>>a>>b;
        ll l=0;
        ll r=1e9;
        ll mid;
        ll ans=-1;
        while(l<=r)
        {
            mid=(l+r)>>1;
            ll a1=a*mid-x;
            ll a2=b*mid-y;
            if(a1>=0&&a2>=0&&(a1<=a2))
            {
                ans=mid;
                r=mid-1;
            }else
            {
                l=mid+1;
            }
        }
        if(ans==-1)
        {
            cout<<-1<<endl;
        }else
        {
            cout<<b*ans-y<<endl;
        }
    }
    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '
');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}