#include <bits/stdc++.h>
using namespace std;
void FIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
void printarr(int arr[], int start, int size) {
  for (int i = start; i < size; i++) {
    cout << arr[i] << " ";
  }
}
void debug(int x) { cout << x << "\n"; }
long long int powMod(long long int b, int p) {
  long long int res = 1ll;
  while (p) {
    if (p & 1) {
      res = res * b % 1000000007;
    }
    b = b * b % 1000000007;
    p /= 2;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
bool check(long long int x, long long int y, long long int p, long long int q,
           long long int t) {
  if (p * t >= x && (q * t - p * t) >= (y - x)) {
    return true;
  }
  return false;
}
void solution() {
  long long int x, y, p, q;
  cin >> x >> y >> p >> q;
  long long int l = -1, r = 1e9;
  if (!check(x, y, p, q, r)) {
    cout << -1 << "\n";
    return;
  }
  while (r - l > 1) {
    long long int mid = l + ((r - l) / 2);
    if (check(x, y, p, q, mid)) {
      r = mid;
    } else {
      l = mid;
    }
  }
  cout << r * q - y << "\n";
  return;
}
int main() {
  FIO();
  int t;
  t = 1;
  cin >> t;
  while (t--) {
    solution();
  }
  return 0;
}

### Prompt

Please formulate a cpp solution 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;
void FIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
void printarr(int arr[], int start, int size) {
  for (int i = start; i < size; i++) {
    cout << arr[i] << " ";
  }
}
void debug(int x) { cout << x << "\n"; }
long long int powMod(long long int b, int p) {
  long long int res = 1ll;
  while (p) {
    if (p & 1) {
      res = res * b % 1000000007;
    }
    b = b * b % 1000000007;
    p /= 2;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
bool check(long long int x, long long int y, long long int p, long long int q,
           long long int t) {
  if (p * t >= x && (q * t - p * t) >= (y - x)) {
    return true;
  }
  return false;
}
void solution() {
  long long int x, y, p, q;
  cin >> x >> y >> p >> q;
  long long int l = -1, r = 1e9;
  if (!check(x, y, p, q, r)) {
    cout << -1 << "\n";
    return;
  }
  while (r - l > 1) {
    long long int mid = l + ((r - l) / 2);
    if (check(x, y, p, q, mid)) {
      r = mid;
    } else {
      l = mid;
    }
  }
  cout << r * q - y << "\n";
  return;
}
int main() {
  FIO();
  int t;
  t = 1;
  cin >> t;
  while (t--) {
    solution();
  }
  return 0;
}

```