Codeforces Round #412 C. Success Rate

You are an experienced Codeforces user. Today you found out that during your activity on Codeforces you have made ysubmissions, 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 to9 / 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.

解题思路：

这道题题意比较好懂，就不说了，为了方便理解我们可以用 (x+a)/(y+b) = p/q 来表示其核心思想，其中a 为做对的题目，b为做的题目，则有x+a = k*p,y+b = k*q.且有0<=a<=b，两式合并可得k>=x/p,k>=(y-x)/(q-p), 则如要满足两个等式，k应取其中较大的值(注意:若取得的k为小数，我们需要的是最小整数解，加1)，最后注意一下特殊情况 p = 0,p = q.

实现代码：

#include <bits/stdc++.h>
using namespace std;

int main() {
    long long n, x, y, p, q, k;
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> x >> y >> p >> q;
        if (p == 0) {
            if (x == 0) cout << 0 << endl;
            else cout << -1 << endl;
        }
        else if (p == q) {
            if (x == y) cout << 0 << endl;
            else cout << -1 << endl;
        }
        else {
                long long  k1 = (y-x)/(q-p) + (((y-x)%(q-p)) ? 1 : 0);
                long long  k2 = x/p + ((x%p) ? 1 : 0);
            k = max(k1,k2);
            cout << k*q-y << endl;
        }
    }
    return 0;
}