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;
long long int power(long long int a, long long int b) {
  long long int x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y);
      if (x > 1000000007) x %= 1000000007;
    }
    y = (y * y);
    if (y > 1000000007) y %= 1000000007;
    b /= 2;
  }
  return x;
}
int main() {
  long double a, b, c;
  long long int l, r, m;
  l = 2;
  r = 1000000000;
  cin >> a >> b;
  cout.precision(12);
  cout.setf(cout.fixed);
  c = a + b;
  if (a == b) {
    cout << a << endl;
    return 0;
  }
  if (b > a) {
    cout << -1 << endl;
    return 0;
  }
  if (c < b) {
    cout << "-1" << endl;
    return 0;
  }
  if (c / 2 < b) {
    cout << c << endl;
    return 0;
  }
  while (l != r) {
    m = (l + r) / 2;
    if (c / 2.0 / m >= b) {
      l = m + 1;
    } else {
      r = m;
    }
  }
  c /= 2.0;
  c /= (long double)(l - 1);
  cout << c << endl;
  return 0;
}