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 double eps = 1e-10;
int a, b;
int sign(double x) {
  if (fabs(x) < eps) return 0;
  return x > 0 ? 1 : -1;
}
bool judge(int dis, long long x) {
  if (x == 0) {
    if (a == b)
      return true;
    else
      return false;
  }
  int y = 1.0 * dis / x;
  if (sign(y - b) < 0) return false;
  return true;
}
long long solve(int dis) {
  long long l = 0, r = (int)1e11;
  long long ret = -1;
  while (l <= r) {
    long long mid = (l + r) >> 1;
    if (judge(dis, mid * 2)) {
      ret = mid;
      l = mid + 1;
    } else
      r = mid - 1;
  }
  return ret * 2;
}
int main() {
  scanf("%d%d", &a, &b);
  long long ret, rem;
  ret = solve(a + b);
  rem = solve(a - b);
  double ans, tmp;
  if (rem == 0)
    tmp = b;
  else
    tmp = 1.0 * (a - b) / rem;
  if (ret < 0 && rem < 0)
    puts("-1");
  else if (ret < 0)
    printf("%.10f\n", 1.0 * (a - b) / rem);
  else if (rem < 0)
    printf("%.10f\n", 1.0 * (a + b) / ret);
  else
    printf("%.10f\n", min(tmp, 1.0 * (a + b) / ret));
  return 0;
}