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;
long long findMaxTwos(long long num) {
  long long count = 0;
  while (num != 0) {
    if (num % 2 == 0) {
      count++;
      num = num / 2;
    } else
      return count;
  }
  return count;
}
double binarySearch(double num, long long start, long long end, long long b) {
  if (start == end) return (double)num / (double)(start);
  long long mid = (start + end) / 2;
  if (num / (double)mid < (double)b)
    return binarySearch(num, start, mid - 1, b);
  if (num / (double)(mid + 1) < b) {
    return num / (double)mid;
  }
  return binarySearch(num, mid + 1, end, b);
}
double solve(double num, long long b) {
  long long start = 2;
  if (double(num) < (double)b) return -1;
  double ans = binarySearch(num, 1, (int)num + 1, b);
  return ans;
}
int main() {
  int t;
  t = 1;
  while (t--) {
    long long a, b;
    scanf("%lld %lld", &a, &b);
    double sol1 = solve((double)(a + b) / (double)2, b);
    double sol2;
    if (a < b)
      sol2 = -1;
    else
      sol2 = solve((double)(a - b) / (double)2, b);
    if (sol1 == -1 && sol2 == -1)
      printf("%d", -1);
    else if (sol1 == -1)
      printf("%.9lf", sol2);
    else if (sol2 == -1)
      printf("%.9lf", sol1);
    else {
      if (sol1 > sol2)
        printf("%.9f", sol2);
      else
        printf("%.9f", sol1);
    }
  }
  return 0;
}