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() {
  double a, b;
  scanf("%lf %lf", &a, &b);
  double two = 2.0;
  double fin1 = 12309131309;
  double fin2 = 12309131309;
  if (a - b >= 0.0) {
    if (a - b == 0) {
      fin1 = a;
    } else {
      double k = (a - b);
      k = k / two;
      k = k / b;
      k = (int)k;
      double t1 = a - b;
      t1 /= two;
      t1 /= k;
      double lo = two * k * t1;
      double hi = lo + t1;
      if (a >= lo && a <= hi) {
        fin1 = t1;
      }
    }
  }
  if (a - b >= 0.0) {
    double k1 = a - b;
    k1 /= two;
    k1 /= b;
    k1 = (int)k1;
    double t2 = a + b;
    t2 /= two;
    t2 /= (double)(k1 + 1.0);
    double lo1 = two * k1 * t2;
    lo1 = lo1 + t2;
    double hi1 = lo1 + t2;
    if (a >= lo1 && a <= hi1) {
      fin2 = t2;
    }
  }
  double ans = min(fin1, fin2);
  if (ans != 12309131309) {
    printf("%0.9lf", ans);
  } else {
    printf("-1\n");
  }
}