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;
bool check(long long x, long long y, long long p, long long q, long long t) {
  return p * t >= x && q * t - p * t >= y - x;
}
void solve() {
  long long x, y, p, q;
  scanf("%lld%lld%lld%lld", &x, &y, &p, &q);
  long long l = -1;
  long long r = (long long)1e9;
  if (!check(x, y, p, q, r)) {
    printf("-1\n");
    return;
  }
  while (r - l > 1) {
    long long m = (l + r) / 2;
    if (check(x, y, p, q, m)) {
      r = m;
    } else {
      l = m;
    }
  }
  printf("%lld\n", r * q - y);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    solve();
  }
  return 0;
}
