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;
int main() {
  int a, b, x, y;
  double ans = 0x7fffffff;
  while (scanf("%d%d", &x, &y) == 2) {
    if (y > x) {
      printf("-1\n");
      continue;
    } else if (y == 0) {
      printf("0.000000000001\n");
      continue;
    }
    b = x + y;
    int k = 1;
    int t;
    ans = 0x7fffffff;
    for (;;) {
      t = k;
      k *= 2;
      if (b * 1.0 / k >= y)
        ans = ans < b * 1.0 / k ? ans : b * 1.0 / k;
      else
        break;
    }
    while (t < k) {
      int mid = (t + k) / 2;
      if (mid % 2 == 1) break;
      if (b * 1.0 / mid >= y)
        t = mid;
      else
        k = mid;
    }
    double s1 = b * 1.0 / k;
    double s2 = b * 1.0 / t;
    if (s1 >= y && s2 >= y)
      ans = s1 > s2 ? s2 : s1;
    else if (s1 >= y)
      ans = s1;
    else
      ans = s2;
    b = y - x;
    if (b == 0) {
      printf("%.12lf\n", (double)y);
      continue;
    }
    k = 0;
    for (;;) {
      k += 2;
      if (b * 1.0 / k >= y)
        ans = ans < b * 1.0 / k ? ans : b * 1.0 / k;
      else
        break;
    }
    printf("%.12lf\n", ans);
  }
  return 0;
}