#include <bits/stdc++.h>
using namespace std;
inline void read(long long &x) {
  char ch;
  bool flag = false;
  for (ch = getchar(); !isdigit(ch); ch = getchar())
    if (ch == '-') flag = true;
  for (x = 0; isdigit(ch); x = x * 10 + ch - '0', ch = getchar())
    ;
  x = flag ? -x : x;
}
inline void read(int &x) {
  char ch;
  bool flag = false;
  for (ch = getchar(); !isdigit(ch); ch = getchar())
    if (ch == '-') flag = true;
  for (x = 0; isdigit(ch); x = x * 10 + ch - '0', ch = getchar())
    ;
  x = flag ? -x : x;
}
inline void write(long long x) {
  static const int maxlen = 100;
  static char s[maxlen];
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (!x) {
    putchar('0');
    return;
  }
  int len = 0;
  for (; x; x /= 10) s[len++] = x % 10 + '0';
  for (int i = len - 1; i >= 0; --i) putchar(s[i]);
}
inline void write(int x) {
  static const int maxlen = 100;
  static char s[maxlen];
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (!x) {
    putchar('0');
    return;
  }
  int len = 0;
  for (; x; x /= 10) s[len++] = x % 10 + '0';
  for (int i = len - 1; i >= 0; --i) putchar(s[i]);
}
int n;
long long x, y, p, q;
long long gcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1ll;
    y = 0ll;
    return a;
  }
  long long g = gcd(b, a % b, x, y);
  long long t = x;
  x = y;
  y = t - (a / b * x);
  return g;
}
void doit() {
  read(x);
  read(y);
  read(p);
  read(q);
  if (p == 0)
    if (x != 0) {
      puts("-1");
      return;
    } else {
      puts("0");
      return;
    }
  long long T1 = p * y - q * x, M1;
  if (T1 == 0) {
    puts("0");
    return;
  }
  M1 = (q - T1 % q) % q;
  long long X1, X2;
  long long g = gcd(p, q, X1, X2);
  if (q == p) {
    puts("-1");
    return;
  }
  if (M1 % g != 0) {
    puts("-1");
    return;
  }
  long long ans = X1 * (M1 / g);
  long long tmp = (q / g);
  long long lim = T1 / (q - p);
  if (ans > lim) {
    ans = ans - (ans - lim) / tmp * tmp;
    int tot = 0;
    while (ans > lim) {
      ans -= tmp;
      tot++;
      if (tot >= 3) return;
    }
    if (ans < lim) ans += tmp;
  } else if (ans < lim) {
    ans = ans + (lim - ans) / tmp * tmp;
    int tot = 0;
    while (ans < lim) {
      ans += tmp;
      tot++;
      if (tot >= 3) return;
    }
  }
  if (T1 <= 0) {
    long long XXX = ans, YYY = T1, ZZZ = p;
    long long TTT = (-T1 + p - 1) / p;
    ans += (TTT - ans + tmp - 1) / tmp * tmp;
    int tot = 0;
    while (T1 + p * ans < 0) {
      ans += tmp;
      tot++;
    }
    if (tot >= 3) printf("%I64d  %I64d  %I64d\n", XXX, YYY, ZZZ);
  }
  printf("%I64d\n", ans);
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++) doit();
  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;
inline void read(long long &x) {
  char ch;
  bool flag = false;
  for (ch = getchar(); !isdigit(ch); ch = getchar())
    if (ch == '-') flag = true;
  for (x = 0; isdigit(ch); x = x * 10 + ch - '0', ch = getchar())
    ;
  x = flag ? -x : x;
}
inline void read(int &x) {
  char ch;
  bool flag = false;
  for (ch = getchar(); !isdigit(ch); ch = getchar())
    if (ch == '-') flag = true;
  for (x = 0; isdigit(ch); x = x * 10 + ch - '0', ch = getchar())
    ;
  x = flag ? -x : x;
}
inline void write(long long x) {
  static const int maxlen = 100;
  static char s[maxlen];
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (!x) {
    putchar('0');
    return;
  }
  int len = 0;
  for (; x; x /= 10) s[len++] = x % 10 + '0';
  for (int i = len - 1; i >= 0; --i) putchar(s[i]);
}
inline void write(int x) {
  static const int maxlen = 100;
  static char s[maxlen];
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (!x) {
    putchar('0');
    return;
  }
  int len = 0;
  for (; x; x /= 10) s[len++] = x % 10 + '0';
  for (int i = len - 1; i >= 0; --i) putchar(s[i]);
}
int n;
long long x, y, p, q;
long long gcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1ll;
    y = 0ll;
    return a;
  }
  long long g = gcd(b, a % b, x, y);
  long long t = x;
  x = y;
  y = t - (a / b * x);
  return g;
}
void doit() {
  read(x);
  read(y);
  read(p);
  read(q);
  if (p == 0)
    if (x != 0) {
      puts("-1");
      return;
    } else {
      puts("0");
      return;
    }
  long long T1 = p * y - q * x, M1;
  if (T1 == 0) {
    puts("0");
    return;
  }
  M1 = (q - T1 % q) % q;
  long long X1, X2;
  long long g = gcd(p, q, X1, X2);
  if (q == p) {
    puts("-1");
    return;
  }
  if (M1 % g != 0) {
    puts("-1");
    return;
  }
  long long ans = X1 * (M1 / g);
  long long tmp = (q / g);
  long long lim = T1 / (q - p);
  if (ans > lim) {
    ans = ans - (ans - lim) / tmp * tmp;
    int tot = 0;
    while (ans > lim) {
      ans -= tmp;
      tot++;
      if (tot >= 3) return;
    }
    if (ans < lim) ans += tmp;
  } else if (ans < lim) {
    ans = ans + (lim - ans) / tmp * tmp;
    int tot = 0;
    while (ans < lim) {
      ans += tmp;
      tot++;
      if (tot >= 3) return;
    }
  }
  if (T1 <= 0) {
    long long XXX = ans, YYY = T1, ZZZ = p;
    long long TTT = (-T1 + p - 1) / p;
    ans += (TTT - ans + tmp - 1) / tmp * tmp;
    int tot = 0;
    while (T1 + p * ans < 0) {
      ans += tmp;
      tot++;
    }
    if (tot >= 3) printf("%I64d  %I64d  %I64d\n", XXX, YYY, ZZZ);
  }
  printf("%I64d\n", ans);
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++) doit();
  return 0;
}

```