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.
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long x, long long y) {
  if (!y) return x;
  return gcd(y, x % y);
}
int main() {
  int tt;
  scanf("%d", &tt);
  while (tt--) {
    long long x, y, a, b;
    scanf("%I64d %I64d %I64d %I64d", &x, &y, &a, &b);
    if (a == b)
      if (x != y)
        printf("-1\n");
      else
        printf("0\n");
    else if (a == 0) {
      if (x == 0)
        printf("0\n");
      else
        printf("-1\n");
    } else {
      long long d = gcd(a, b);
      long long now1 = a / d;
      long long now2 = b / d;
      long long k1, k2;
      k1 = x / now1;
      if (x % now1) k1++;
      k2 = (y - x) / (now2 - now1);
      if ((y - x) % (now2 - now1)) k2++;
      printf("%I64d\n", max(d, max(k1, k2)) * now2 - y);
    }
  }
  return 0;
}
