Codeforces Round #412 C. Success Rate

Success Rate

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.

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

#include <bits/stdc++.h>
using namespace std;
#define ll long long
ll x, y, q, p;
int main(){
    int t;
    cin >> t;
    while(t--){
        cin >> x >> y >> p >> q;
        if(q == p){
            printf("%d
",(x==y)?0:-1);
            continue;
        }
        if(p == 0){
            printf("%d
",(x==0)?0:-1);
            continue;
        }
        ll GCD = __gcd(q,p);
        q/=GCD; p/=GCD;
        ll n = max(ceil(x*1.0/p),ceil((y-x)*1.0/(q-p)));
        printf("%lld
",n*q-y);
    }
    return 0;
}