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;
int x, y, mn, t1, t2;
bool f1, f2;
double k1, k2;
inline double ABS(double x) { return x > 0 ? x : -x; }
int main() {
  scanf("%d%d", &x, &y);
  if (y > x) {
    puts("-1");
    return 0;
  }
  if (y == x) {
    k1 = x * 1.0;
    printf("%.9lf\n", k1);
    return 0;
  }
  mn = y << 1;
  t1 = x - y;
  t2 = t1 / mn;
  if (t2) {
    k1 = (1.0 * t1) / (1.0 * t2) / 2.0;
    f1 = 1;
  }
  t1 = x + y;
  t2 = t1 / mn;
  if (t2) {
    f2 = 1;
    k2 = (1.0 * t1) / (1.0 * t2) / 2.0;
  }
  if (f1 && f2) {
    if (k1 < k2)
      printf("%.9lf\n", k1);
    else
      printf("%.9lf\n", k2);
  } else if (f1)
    printf("%.9lf\n", k1);
  else if (f2)
    printf("%.9lf\n", k2);
  else
    puts("-1");
  return 0;
}
