#include <bits/stdc++.h>
using namespace std;
const unsigned mod = 998244353;
const unsigned _mod = 1e9 + 7;
const long long infi = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
int read() {
  int x = 0;
  char ch = getchar();
  while (ch > '9' || ch < '0') ch = getchar();
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x;
}
int read2() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long lread() {
  long long x = 0;
  char ch = getchar();
  while (ch > '9' || ch < '0') ch = getchar();
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x;
}
inline long long ksm(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y & 1ll) ret = ret * x % mod, y--;
    y >>= 1ll;
    x = x * x % mod;
  }
  return ret;
}
inline int fmax(int x, int y) { return x > y ? x : y; }
inline int fmin(int x, int y) { return x < y ? x : y; }
inline long long gmax(long long x, long long y) { return x > y ? x : y; }
inline long long gmin(long long x, long long y) { return x < y ? x : y; }
inline void fswap(int &x, int &y) { x ^= y ^= x ^= y; }
inline void gswap(long long &x, long long &y) { x ^= y ^= x ^= y; }
inline long long _gcd(long long x, long long y) {
  return y ? _gcd(y, x % y) : x;
}
long long T, x, y, p, q, a, z;
int main() {
  cin >> T;
  while (T--) {
    x = lread(), y = lread(), p = lread(), q = lread();
    if (q == p && y != x || !p && x)
      a = -1;
    else if (q == p && y == x || !p && !x)
      a = 0;
    else {
      y -= x, q -= p;
      z = gmax((x + p - 1) / p, (y + q - 1) / q);
      a = q * z - y + p * z - x;
    }
    printf("%lld\n", a);
  }
  return 0;
}

### Prompt

Generate a CPP solution 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;
const unsigned mod = 998244353;
const unsigned _mod = 1e9 + 7;
const long long infi = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
int read() {
  int x = 0;
  char ch = getchar();
  while (ch > '9' || ch < '0') ch = getchar();
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x;
}
int read2() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long lread() {
  long long x = 0;
  char ch = getchar();
  while (ch > '9' || ch < '0') ch = getchar();
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x;
}
inline long long ksm(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y & 1ll) ret = ret * x % mod, y--;
    y >>= 1ll;
    x = x * x % mod;
  }
  return ret;
}
inline int fmax(int x, int y) { return x > y ? x : y; }
inline int fmin(int x, int y) { return x < y ? x : y; }
inline long long gmax(long long x, long long y) { return x > y ? x : y; }
inline long long gmin(long long x, long long y) { return x < y ? x : y; }
inline void fswap(int &x, int &y) { x ^= y ^= x ^= y; }
inline void gswap(long long &x, long long &y) { x ^= y ^= x ^= y; }
inline long long _gcd(long long x, long long y) {
  return y ? _gcd(y, x % y) : x;
}
long long T, x, y, p, q, a, z;
int main() {
  cin >> T;
  while (T--) {
    x = lread(), y = lread(), p = lread(), q = lread();
    if (q == p && y != x || !p && x)
      a = -1;
    else if (q == p && y == x || !p && !x)
      a = 0;
    else {
      y -= x, q -= p;
      z = gmax((x + p - 1) / p, (y + q - 1) / q);
      a = q * z - y + p * z - x;
    }
    printf("%lld\n", a);
  }
  return 0;
}

```