#include <bits/stdc++.h>
using namespace std;
using ll = long long;
ll fit(ll y1, ll x, ll y, ll p, ll q) {
  ll dx = p * (y1 / q) - x;
  return 0 <= dx && dx <= y1 - y;
}
ll solve(int x, int y, int p, int q) {
  if (x > 0 && p == 0 || p == 1 && q == 1 && (x % y != 0 || x / y != 1)) {
    return -1;
  }
  ll y1 = (y % q == 0 ? y : y + q - y % q), y2 = y1 + q, i = 1;
  while (!fit(y2, x, y, p, q)) {
    i <<= 1;
    y2 = y1 + i * q;
  }
  if (fit(y1, x, y, p, q)) {
    return y1 - y;
  }
  ll m, a = 0, b = i;
  while (a + 1 < b) {
    m = (a + b) / 2;
    if (!fit(y1 + m * q, x, y, p, q)) {
      a = m;
    } else {
      b = m;
    }
  }
  return y1 + b * q - y;
}
int main() {
  ios_base::sync_with_stdio(false);
  int t, x, y, p, q;
  cin >> t;
  while (t-- > 0) {
    cin >> x >> y >> p >> q;
    cout << solve(x, y, p, q) << "\n";
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
using ll = long long;
ll fit(ll y1, ll x, ll y, ll p, ll q) {
  ll dx = p * (y1 / q) - x;
  return 0 <= dx && dx <= y1 - y;
}
ll solve(int x, int y, int p, int q) {
  if (x > 0 && p == 0 || p == 1 && q == 1 && (x % y != 0 || x / y != 1)) {
    return -1;
  }
  ll y1 = (y % q == 0 ? y : y + q - y % q), y2 = y1 + q, i = 1;
  while (!fit(y2, x, y, p, q)) {
    i <<= 1;
    y2 = y1 + i * q;
  }
  if (fit(y1, x, y, p, q)) {
    return y1 - y;
  }
  ll m, a = 0, b = i;
  while (a + 1 < b) {
    m = (a + b) / 2;
    if (!fit(y1 + m * q, x, y, p, q)) {
      a = m;
    } else {
      b = m;
    }
  }
  return y1 + b * q - y;
}
int main() {
  ios_base::sync_with_stdio(false);
  int t, x, y, p, q;
  cin >> t;
  while (t-- > 0) {
    cin >> x >> y >> p >> q;
    cout << solve(x, y, p, q) << "\n";
  }
  return 0;
}

```