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>
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1005;
int main() {
  int a, b;
  scanf("%d%d", &a, &b);
  double res = -1;
  if (a - b > 0) {
    int c = a - b;
    int x = c / b;
    if (x % 2 != 0) {
      x--;
    }
    if (x >= 2) {
      res = 1.0 * c / x;
    }
  } else if (a == b) {
    if (res == -1) {
      res = a;
    } else {
      res = min(1.0 * a, res);
    }
  }
  int d = a + b;
  int x = d / b;
  if (x % 2 != 0) {
    x--;
  }
  if (x >= 2) {
    if (res == -1) {
      res = 1.0 * d / x;
    } else {
      res = min(res, 1.0 * d / x);
    }
  }
  if (res == -1) {
    printf("-1\n");
  } else {
    printf("%.12lf\n", res);
  }
  return 0;
}
