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;
const int INFINITE = 1000000010;
int main(int argc, char *argv[]) {
  int a, b;
  cin >> a >> b;
  if (a < b) {
    std::cout << -1 << std::endl;
    return 0;
  }
  if (a == b) {
    std::cout << a << std::endl;
    return 0;
  }
  double case1, case2;
  int dividor;
  if ((a - b) / b == 0) {
    case1 = INFINITE;
  } else {
    dividor = (a - b) / b;
    if (dividor % 2 != 0) dividor--;
    if (dividor == 0)
      case1 = INFINITE;
    else
      case1 = ((double)(a - b)) / dividor;
  }
  dividor = (a - b) / b;
  if (dividor % 2 != 0) dividor--;
  case2 = ((double)(a + b)) / (dividor + 2);
  cout.precision(10);
  std::cout << ((case1 < case2) ? case1 : case2) << std::endl;
  return 0;
}