Problem: The determinant of a matrix 2 × 2 is defined as follows:

<image>

A matrix is called degenerate if its determinant is equal to zero. 

The norm ||A|| of a matrix A is defined as a maximum of absolute values of its elements.

You are given a matrix <image>. Consider any degenerate matrix B such that norm ||A - B|| is minimum possible. Determine ||A - B||.

Input

The first line contains two integers a and b (|a|, |b| ≤ 109), the elements of the first row of matrix A. 

The second line contains two integers c and d (|c|, |d| ≤ 109) the elements of the second row of matrix A.

Output

Output a single real number, the minimum possible value of ||A - B||. Your answer is considered to be correct if its absolute or relative error does not exceed 10 - 9.

Examples

Input

1 2
3 4


Output

0.2000000000


Input

1 0
0 1


Output

0.5000000000

Note

In the first sample matrix B is <image>

In the second sample matrix B is <image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
int IntMaxVal = (int)1e20;
int IntMinVal = (int)-1e20;
long long LongMaxVal = (long long)1e20;
long long LongMinVal = (long long)-1e20;
template <typename T>
inline void minimize(T& a, T b) {
  a = std::min(a, b);
}
template <typename T>
inline void maximize(T& a, T b) {
  a = std::max(a, b);
}
using namespace std;
template <typename T>
struct argument_type;
template <typename T, typename U>
struct argument_type<T(U)> {};
template <typename T1, typename T2>
istream& operator>>(istream& is, pair<T1, T2>& s) {
  is >> s.first >> s.second;
  return is;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  for (int i = 0; i < v.size(); i++) os << v[i] << ' ';
  os << '\n';
  return os;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& s, const pair<T1, T2>& t) {
  s << t.first << ' ' << t.second;
  return s;
}
template <typename T>
vector<T> readVector(int n) {
  vector<T> res(n);
  for (int i = 0; i < n; i++) cin >> res[i];
  return res;
}
long long a, b, c, d;
double solve(long long a, long long b, long long c, long long d, int am, int bm,
             int cm, int dm) {
  double ans = 1e100;
  double num = b * c - a * d;
  double den = a * dm + d * am - b * cm - c * bm;
  minimize(ans, 1.0 * max(min(abs(a), abs(d)), min(abs(b), abs(c))));
  if (num == 0)
    ans = 0;
  else if (den != 0) {
    double k = num / den;
    auto den1 = (a + am * k) * (d + dm * k);
    auto den2 = (b + bm * k) * (c + cm * k);
    minimize(ans, fabs(num / den));
  }
  return ans;
}
int main() {
  srand(time(NULL));
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> a >> b >> c >> d;
  double ans = 1e200;
  minimize(ans, solve(a, b, c, d, 1, -1, 1, -1));
  minimize(ans, solve(a, b, c, d, 1, 1, -1, -1));
  minimize(ans, solve(a, b, c, d, 1, 1, 1, 1));
  minimize(ans, solve(a, b, c, d, 1, -1, -1, 1));
  fixed(cout);
  cout << setprecision(10) << ans;
}