Problem: There is a polyline going through points (0, 0) – (x, x) – (2x, 0) – (3x, x) – (4x, 0) – ... - (2kx, 0) – (2kx + x, x) – .... 

We know that the polyline passes through the point (a, b). Find minimum positive value x such that it is true or determine that there is no such x.

Input

Only one line containing two positive integers a and b (1 ≤ a, b ≤ 109).

Output

Output the only line containing the answer. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 9. If there is no such x then output  - 1 as the answer.

Examples

Input

3 1


Output

1.000000000000


Input

1 3


Output

-1


Input

4 1


Output

1.250000000000

Note

You can see following graphs for sample 1 and sample 3. 

<image> <image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline T abs(T t) {
  return t < 0 ? -t : t;
}
const long long modn = 1000000007;
inline long long mod(long long x) { return x % modn; }
int main() {
  int a, b;
  scanf("%d %d", &a, &b);
  if (a == b) {
    printf("%d\n", b);
    return 0;
  }
  bool ok = false;
  double x = 1 / 0.;
  double c = a - b;
  if (c > 0) {
    int l = 1, r = (a + b + 2) / 2;
    while (l < r) {
      int m = (l + r + 1) / 2;
      if ((c / (2. * m)) >= b)
        l = m;
      else
        r = m - 1;
    }
    if (l == r && (c / (2. * l)) >= b) x = c / (2. * l), ok = true;
  }
  c = a + b;
  int l = 1, r = (a + b + 2) / 2;
  while (l < r) {
    int m = (l + r + 1) / 2;
    if ((c / (2. * m)) >= b)
      l = m;
    else
      r = m - 1;
  }
  if (l == r && (c / (2. * l)) >= b) x = min(x, c / (2. * l)), ok = true;
  if (ok)
    printf("%.10f\n", x);
  else
    puts("-1");
}