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;
int a, b;
double ans = 1e18;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> a >> b;
  if (b > a) {
    cout << "-1\n";
    return 0;
  }
  int l = 1, r = 1e9, m = -1;
  while (l <= r) {
    int k = (l + r) / 2;
    double x = (a - b) / (2.0 * k);
    if (x < b) {
      r = k - 1;
    } else {
      m = k;
      l = k + 1;
    }
  }
  if (m != -1) {
    ans = min(ans, (a - b) / (2.0 * m));
  }
  l = 0, r = 1e9, m = -1;
  while (l <= r) {
    int k = (l + r) / 2;
    double x = (a + b) / (2 * k + 2.0);
    if (x < b) {
      r = k - 1;
    } else {
      m = k;
      l = k + 1;
    }
  }
  if (m != -1) {
    ans = min(ans, (a + b) / (2.0 * m + 2.0));
  }
  printf("%.12f\n", ans);
  return 0;
}