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;
template <class T>
inline void smax(T &x, T y) {
  x = max((x), (y));
}
template <class T>
inline void smin(T &x, T y) {
  x = min((x), (y));
}
const int MOD = 1e9 + 7;
const int SIZE = 1e3 + 10;
const int INF = 0x3f3f3f3f;
int a, b;
double getx(int foo) {
  return ((double)a + (foo & 1 ? -b : b)) / (2 * (foo / 2 + 1));
}
int main() {
  if (0) {
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
  }
  while (scanf("%d%d", &(a), &(b)) == 2) {
    if (a == b) {
      printf("%.12f\n", (double)a);
      continue;
    }
    int l = 0, r = INF;
    int mid;
    while (l < r) {
      mid = l + (r - l + 1) / 2;
      if (getx(mid) < b)
        r = mid - 1;
      else
        l = mid;
    }
    if (getx(l) < b) {
      printf("-1\n");
      continue;
    }
    printf("%.12f\n", getx(l));
  }
  return 0;
}