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>
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
using namespace std;
void itval(istream_iterator<string> it) {}
template <typename T, typename... Args>
void itval(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << endl;
  itval(++it, args...);
}
const long long int MOD = 1e9 + 9;
template <typename T>
inline void print(T x) {
  cout << x << "\n";
}
template <typename T>
inline void printvec(T x) {
  for (auto a : x) cout << a << ' ';
  cout << '\n';
}
struct custom {
  bool operator()(const pair<int, pair<int, int> > &p1,
                  const pair<int, pair<int, int> > &p2) const {
    if (p1.first == p2.first) return p1.second.first > p2.second.first;
    return p1.first > p2.first;
  }
};
long long int get_pow(long long int a, long long int b, long long int M = MOD) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * a) % M;
    a = (a * a) % M;
    b >>= 1;
  }
  return res;
}
const long long int N = 1e7 + 5, inf = 2e18;
pair<long double, long double> calc(long double a, long double b, long double c,
                                    long double d) {
  std::vector<long double> v = {a * c, a * d, b * c, b * d};
  sort(v.begin(), v.end());
  return {v[0], v[3]};
}
void solve() {
  long long int a, b, c, d;
  cin >> a >> b >> c >> d;
  long double lo = 0, hi = 1e12;
  for (long long int i = (long long int)0; i < (long long int)(200); i++) {
    long double mid = (hi + lo) * 0.5;
    auto p1 = calc(a - mid, a + mid, d + mid, d - mid);
    auto p2 = calc(b - mid, b + mid, c - mid, c + mid);
    if (p1.second < p2.first || p2.second < p1.first)
      lo = mid;
    else
      hi = mid;
  }
  cout << fixed << setprecision(12);
  cout << hi << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int test = 1;
  clock_t z = clock();
  for (long long int tes = (long long int)0; tes < (long long int)(test);
       tes++) {
    solve();
  }
  fprintf(stderr, "Total Time:%.4f\n", (double)(clock() - z) / CLOCKS_PER_SEC),
      fflush(stderr);
  return 0;
}
