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;
template <typename A, typename B>
int print_value(A title, B val) {
  cout << title << " = " << val << "\n";
  return 1;
}
int break_point() {
  char c;
  while ((c = getchar()) != '\n')
    ;
  return 0;
}
template <typename A>
int logg(A v) {
  cout << v << "\n";
  return 0;
}
void read_intq(int &r) {
  r = 0;
  char c;
  while (1) {
    c = getchar();
    if (c != ' ' && c != '\n') {
      r = c - '0';
      break;
    }
  }
  while (1) {
    c = getchar();
    if (c == ' ' || c == '\n') return;
    r = r * 10 + (c - '0');
  }
}
void read_llq(long long &r) {
  r = 0;
  char c;
  while (1) {
    c = getchar();
    if (c != ' ' && c != '\n') {
      r = c - '0';
      break;
    }
  }
  while (1) {
    c = getchar();
    if (c == ' ' || c == '\n') return;
    r = r * 10 + (c - '0');
  }
}
void run();
int main() {
  srand(time(NULL));
  do {
    run();
    if (0) {
      0 ? printf("-------------------------------\n") : 0;
      0 ? printf("-------------------------------\n") : 0;
    }
  } while (0);
  return 0;
}
void run() {
  long long a, b;
  scanf("%lld%lld", &a, &b);
  if (a < b) {
    printf("-1\n");
    return;
  } else {
    long long c = a + b;
    long long d1 = c / b;
    if (d1 & 1) d1--;
    c = a - b;
    long long d2 = c / b;
    if (d2 & 1) d2--;
    0 ? print_value("double(a + b) / d1", double(a + b) / d1) : 0;
    0 ? print_value("double(a - b) / d2", double(a - b) / d2) : 0;
    printf("%.*lf", 12, min(double(a + b) / d1, double(a - b) / d2));
    putchar('\n');
  }
}
