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 long long inf = (long long)(1e17);
const double E = double(1e-11);
int main() {
  double x, y;
  cin >> x >> y;
  bool f1 = 1, f2 = 1;
  double b1 = x + y;
  if (y > (b1 / 2.0 + E)) {
    f1 = 0;
  }
  double b2 = x - y;
  if ((b2 < 0 + E) || (y > (b2 / 2.0 + E))) {
    f2 = 0;
  }
  double ans1 = 10000000000.0, ans2 = 10000000000.0;
  if ((!f1) && (!f2)) {
    cout << -1;
    return 0;
  }
  if (f1) {
    long long l = 1, r = inf;
    while (r - l > 1) {
      long long m = (l + r) / 2;
      if (b1 / (2.0 * m) > y - E) {
        l = m;
      } else {
        r = m;
      }
    }
    ans1 = b1 / (2.0 * l);
  }
  if (f2) {
    long long l = 1, r = inf;
    while (r - l > 1) {
      long long m = (l + r) / 2;
      if (b2 / (2.0 * m) > y - E) {
        l = m;
      } else {
        r = m;
      }
    }
    ans2 = b2 / (2.0 * l);
  }
  if (abs(b2 - 0.0) < E) {
    ans2 = y;
  }
  if (ans1 > ans2 + E) {
    printf("%.10f", ans2);
  } else {
    printf("%.10f", ans1);
  }
  return 0;
}