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 double AE = 1e-13;
bool ok(int m1, int m2, double x1, double x2) {
  double p1, p2;
  long long t1, t2;
  p1 = (m1 + m2) / (2 * x1);
  p2 = (m1 + m2) / (2 * x2);
  t1 = p1 = floor(p1 + AE);
  t2 = p2 = ceil(p2);
  if (t1 >= t2) return true;
  p1 = (m1 - m2) / (2 * x1);
  p2 = (m1 - m2) / (2 * x2);
  t1 = p1 = floor(p1 + AE);
  t2 = p2 = ceil(p2);
  if (t1 >= t2) return true;
  return false;
}
double Find(int m1, int m2) {
  double ret = 1e19, down = m2, up = 1e18, mid;
  for (int i = 0; i < 100000; i++) {
    mid = (down + up) / 2.0;
    if (ok(m1, m2, down, mid))
      up = mid;
    else
      down = mid;
  }
  return mid;
}
int main() {
  int m1, m2;
  scanf("%d%d", &m1, &m2);
  if (m1 < m2) {
    printf("-1\n");
    return 0;
  }
  double ans = Find(m1, m2);
  if (ans > 1e17) {
    printf("-1\n");
    return 0;
  }
  printf("%.13lf\n", ans);
  return 0;
}
