PROBLEMSET里面神tm搜不到这题，很迷。所以标题就只好注明比赛出处而没法标明题号了。

Description

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.

Key

给你参加的总场次 y 与其中胜利的场次x，现要求胜率达到 p/q ，问还要再比多少场（每一场可胜可负）。 想明白了其实是个巨简单的题目。为了达到目标，比赛的总场次必为 q 的倍数（现设倍数为k），则 k∗q>=y ；在比了 k∗q 场次的情况下，必须胜利 k∗p 次，则 k∗p>=x 。胜利的场数肯定不大于总场数，则只要找到 k∗p−x<=k∗q−y 的最小k即获得了答案。 综合一下三个条件，找出满足以下条件的k的最小值：

k>=y/q; k>=x/p; k>=(y-x)/(q-p);

即 k=max(y/q,x/p,(y−x)/(q−p)) 。其中除法结果若为小数则向上取整。 因此 O(1) 时间复杂度即可求出。 至于无法达到的情况，只有 x/y 不为0而 p/q 为0、或 x/y 不为1而 p/q 为1两种。一开始判断掉就行。

Code

#include<iostream> using namespace std; typedef long long LL; LL T, x, y, p, q; int main() { cin >> T; while (T--) { cin >> x >> y >> p >> q; if (x*q == p*y) { cout << "0\n"; continue; } if (p == 0 || p == q) { cout << "-1\n"; continue; } LL k = (y - x - 1) / (q - p) +1; if (k*p < x) k = (x - 1) / p + 1; if (k*q < y) k = (y - 1) / q + 1; LL res = k*q - y; cout << res << '\n'; } return 0; }