Codeforces Round #412 C. Success Rate

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 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

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 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;
}