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 = 1.2e9;
long long x, y, p, q, r;
inline bool check(long long t) {
  long long m = t * q + r;
  long long x1 = x, x2 = x + m;
  long long z = y + m;
  long long pk = (z / q) * p;
  return pk >= x1 && pk <= x2;
}
int main() {
  int T;
  scanf("%d", &T);
  for (int t = 0; t < (T); ++t) {
    scanf(
        "%lld"
        "%lld"
        "%lld"
        "%lld",
        &x, &y, &p, &q);
    r = q - y % q;
    if (r == q) r = 0;
    if (p == q) {
      if (x == y)
        puts("0");
      else
        puts("-1");
    } else if (p == 0) {
      if (x == 0)
        puts("0");
      else
        puts("-1");
    } else {
      long long L = -1, R = INF;
      while (R - L > 1) {
        long long M = (L + R) / 2;
        if (check(M))
          R = M;
        else
          L = M;
      }
      if (!check(R))
        puts("-1");
      else
        printf(
            "%lld"
            "\n",
            R * q + r);
    }
  }
  return 0;
}
