#include <bits/stdc++.h>
using namespace std;
inline long long get() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
bool check(long long x, long long y, long long k, long long kk) {
  if (x < k || y < kk) return 0;
  return 1;
}
bool check1(long long x, long long y, long long k, long long kk) {
  if (kk + x - k > y) return 0;
  return 1;
}
int main() {
  long long T = get();
  while (T--) {
    long long x, y, xx, yy;
    x = get();
    y = get();
    xx = get();
    yy = get();
    long long l = 1, r = 1ll << 30, tp = -1;
    while (l <= r) {
      long long mid = (l + r) / 2;
      if (check(mid * xx, mid * yy, x, y))
        tp = mid, r = mid - 1;
      else
        l = mid + 1;
    }
    l = tp;
    r = 1ll << 30;
    tp = -1;
    while (l <= r) {
      long long mid = (l + r) / 2;
      if (check1(mid * xx, mid * yy, x, y))
        tp = mid, r = mid - 1;
      else
        l = mid + 1;
    }
    if (tp == -1 || tp * xx - x + tp * yy - y - (tp * xx - x) < 0)
      cout << "-1\n";
    else
      cout << tp * xx - x + tp * yy - y - (tp * xx - x) << "\n";
  }
  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;
inline long long get() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
bool check(long long x, long long y, long long k, long long kk) {
  if (x < k || y < kk) return 0;
  return 1;
}
bool check1(long long x, long long y, long long k, long long kk) {
  if (kk + x - k > y) return 0;
  return 1;
}
int main() {
  long long T = get();
  while (T--) {
    long long x, y, xx, yy;
    x = get();
    y = get();
    xx = get();
    yy = get();
    long long l = 1, r = 1ll << 30, tp = -1;
    while (l <= r) {
      long long mid = (l + r) / 2;
      if (check(mid * xx, mid * yy, x, y))
        tp = mid, r = mid - 1;
      else
        l = mid + 1;
    }
    l = tp;
    r = 1ll << 30;
    tp = -1;
    while (l <= r) {
      long long mid = (l + r) / 2;
      if (check1(mid * xx, mid * yy, x, y))
        tp = mid, r = mid - 1;
      else
        l = mid + 1;
    }
    if (tp == -1 || tp * xx - x + tp * yy - y - (tp * xx - x) < 0)
      cout << "-1\n";
    else
      cout << tp * xx - x + tp * yy - y - (tp * xx - x) << "\n";
  }
  return 0;
}

```