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 main() {
  int a, b;
  cin >> a >> b;
  if (a == b)
    cout << a << endl;
  else {
    double res;
    double a1 = a, b1 = b;
    int k1 = -1, k2 = -1;
    double x1, x2;
    if (a > b) {
      res = (a1 - b1) / 2;
      int left = 1;
      int right = a + 1;
      while (1) {
        if (left == right) break;
        int mid = (left + right) / 2;
        if (res / double(mid) < double(b))
          right = mid;
        else if (res / double(mid) >= double(b))
          k1 = mid, x1 = res / double(mid), left = mid + 1;
      }
    }
    int left = 1;
    int right = a + 1;
    res = (a1 + b1) / 2;
    while (1) {
      if (left == right) break;
      int mid = (left + right) / 2;
      if (res / double(mid) < double(b))
        right = mid;
      else if (res / double(mid) >= double(b))
        k2 = mid, x2 = res / double(mid), left = mid + 1;
    }
    if (k1 == -1 && k2 == -1)
      cout << "-1" << endl;
    else if (k1 == -1)
      printf("%.9f", x2);
    else if (k2 == -1)
      printf("%.9f", x1);
    else if (x1 < x2)
      printf("%.9f\n", x1);
    else
      printf("%.9f\n", x2);
  }
  return 0;
}