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;
const long long inf = 1000000007;
const double eps = 1e-12;
const int maxn = 1e4 + 7;
long long n, m, d, k, x, y, p, q;
int main() {
  int t;
  cin >> t;
  while (t--) {
    scanf("%lld%lld%lld%lld", &x, &y, &p, &q);
    long long l = 1, r = 1e18 / q;
    long long ans;
    while (1) {
      m = (l + r) >> 1;
      if (l == m) {
        long long del = l * q - y;
        if (del >= 0 && (x + del) >= l * p && x <= l * p)
          ans = l;
        else if ((r * q - y) >= 0 && (x + (r * q - y)) >= r * p && x <= r * p)
          ans = r;
        else
          ans = -1;
        break;
      }
      if (m * q < y) {
        l = m + 1;
        continue;
      }
      long long del = m * q - y;
      if ((x + del) >= m * p && x <= m * p)
        r = m;
      else
        l = m + 1;
    }
    printf("%lld\n", ans == -1 ? -1 : ans * q - y);
  }
}
