#include <bits/stdc++.h>
using namespace std;
long long t, x, y, p, q, ans;
long double aim;
bool check(long long e) {
  if (e < 0) return 0;
  return (long double)x * q <= 0.5 + (long double)(y + e) * p &&
         (long double)p * (y + e) <= 0.5 + (long double)q * (x + e);
}
int main() {
  for (cin >> t; t--;) {
    cin >> x >> y >> p >> q;
    if (p == q) {
      if (x == y)
        puts("0");
      else
        puts("-1");
      continue;
    }
    if (p == 0) {
      if (x == 0)
        puts("0");
      else
        puts("-1");
      continue;
    }
    aim = max(((long double)x * q - p * y) / (long double)p,
              ((long double)p * y - x * q) / (long double)(q - p));
    ans = aim - 0.5;
    ans = (y + ans + q - 1) / q * q - y;
    while (!check(ans)) ans += q;
    printf("%lld\n", ans);
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long t, x, y, p, q, ans;
long double aim;
bool check(long long e) {
  if (e < 0) return 0;
  return (long double)x * q <= 0.5 + (long double)(y + e) * p &&
         (long double)p * (y + e) <= 0.5 + (long double)q * (x + e);
}
int main() {
  for (cin >> t; t--;) {
    cin >> x >> y >> p >> q;
    if (p == q) {
      if (x == y)
        puts("0");
      else
        puts("-1");
      continue;
    }
    if (p == 0) {
      if (x == 0)
        puts("0");
      else
        puts("-1");
      continue;
    }
    aim = max(((long double)x * q - p * y) / (long double)p,
              ((long double)p * y - x * q) / (long double)(q - p));
    ans = aim - 0.5;
    ans = (y + ans + q - 1) / q * q - y;
    while (!check(ans)) ans += q;
    printf("%lld\n", ans);
  }
}

```