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;
void sc(int& a) { scanf("%d", &a); }
void sc(long long int& a) { scanf("%lld", &a); }
void sc(int& a, int& b) {
  sc(a);
  sc(b);
}
void sc(long long int& a, long long int& b) {
  sc(a);
  sc(b);
}
void sc(int& a, int& b, int& c) {
  sc(a, b);
  sc(c);
}
void sc(long long int& a, long long int& b, long long int& c) {
  sc(a, b);
  sc(c);
}
void prl(int a) { printf("%d\n", a); }
void prl(long long int a) { printf("%lld\n", a); }
void prs(int a) { printf("%d ", a); }
void prs(long long int a) { printf("%lld ", a); }
int mod = 1000000007;
long long int modpow(long long int a, long long int b, long long int mod) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b = b / 2;
  }
  return res % mod;
}
int main() {
  int a, b;
  double x1 = LLONG_MAX, x2 = LLONG_MAX, ans = LLONG_MAX;
  int flag = 0;
  sc(a, b);
  if (a == b) {
    printf("%.10lf\n", double(a));
    return 0;
  }
  if (a < b) {
    printf("-1\n");
    return 0;
  }
  long long int low = 0, high = (a - b) / (2 * b), mid;
  if (high > 0) x1 = 1.0 * (a - b) / (2 * high);
  high = (a + b) / (2 * b);
  if (high > 0) x2 = 1.0 * (a + b) / (2 * high);
  if (min(x1, x2) == LLONG_MAX) {
    printf("-1\n");
  } else
    printf("%.10lf\n", min(x1, x2));
}
