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;
class node {
 public:
  long long int second;
  long long int length;
};
void FastIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
}
int main() {
  FastIO();
  long long int t;
  long long int a, b;
  long long int i;
  double mini = 9999999999;
  cin >> a >> b;
  long long int z = a + b;
  long long int y = a - b;
  long long int r = z / b;
  long long int x = y / b;
  if (r * b < z) {
    if (r % 2 == 1) r--;
  } else if (r * b == z) {
    if (r % 2 == 1) r--;
  }
  if (x * b < y) {
    if (x % 2 == 1) x--;
  } else if (x * b == y) {
    if (x % 2 == 1) x--;
  }
  if ((double)z / r >= b) mini = min(mini, (double)z / r);
  if ((double)y / x >= b) mini = min(mini, (double)y / x);
  if (mini == 9999999999)
    printf("-1\n");
  else
    printf("%.12lf\n", mini);
  return 0;
}