#include <bits/stdc++.h>
using namespace std;
long long int tc, x, y, p, q, Ti, t1, t2, n1, n2, l, r, mid, temp, re;
bool check(long long int m) {
  temp = m * q - y;
  if ((x + temp) >= m * p && m * p >= x)
    return 1;
  else
    return 0;
}
long long int gcd(long long int a, long long int b) {
  if (a * b == 0)
    return a + b;
  else
    return gcd(b, a % b);
}
int main() {
  cin >> tc;
  while (tc--) {
    cin >> x >> y >> p >> q;
    if (p == 0 && x != 0) {
      cout << -1 << endl;
      continue;
    } else if (p == 0 && x == 0) {
      cout << 0 << endl;
      continue;
    }
    n1 = x % p;
    n2 = q % p;
    t1 = x / p;
    t2 = y / q;
    if (t1 > t2) {
      re = t1;
      if (x % p) re++;
    } else if (t2 > t1) {
      re = t2;
      if (y % q) re++;
    } else {
      re = t1;
      if (x % p || y % q) re++;
    }
    Ti = -1, l = re, r = 1e9;
    while (l <= r) {
      mid = (l + r) / 2;
      if (check(mid))
        Ti = mid * q - y, r = mid - 1;
      else
        l = mid + 1;
    }
    cout << Ti << endl;
  }
  return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

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 int tc, x, y, p, q, Ti, t1, t2, n1, n2, l, r, mid, temp, re;
bool check(long long int m) {
  temp = m * q - y;
  if ((x + temp) >= m * p && m * p >= x)
    return 1;
  else
    return 0;
}
long long int gcd(long long int a, long long int b) {
  if (a * b == 0)
    return a + b;
  else
    return gcd(b, a % b);
}
int main() {
  cin >> tc;
  while (tc--) {
    cin >> x >> y >> p >> q;
    if (p == 0 && x != 0) {
      cout << -1 << endl;
      continue;
    } else if (p == 0 && x == 0) {
      cout << 0 << endl;
      continue;
    }
    n1 = x % p;
    n2 = q % p;
    t1 = x / p;
    t2 = y / q;
    if (t1 > t2) {
      re = t1;
      if (x % p) re++;
    } else if (t2 > t1) {
      re = t2;
      if (y % q) re++;
    } else {
      re = t1;
      if (x % p || y % q) re++;
    }
    Ti = -1, l = re, r = 1e9;
    while (l <= r) {
      mid = (l + r) / 2;
      if (check(mid))
        Ti = mid * q - y, r = mid - 1;
      else
        l = mid + 1;
    }
    cout << Ti << endl;
  }
  return 0;
}

```