Problem: Problem

You brought a flat, holeless donut with a $ W $ horizontal $ H $ vertical $ H $ rectangle for ACPC.
Place this donut on the $ 2 $ dimension plane coordinate $ (0,0) $ with the center of the donut so that the side of length H and the $ y $ axis are parallel.


On the ACPC $ 1 $ day you ate a donut that was in the range of $ w $ vertical $ h $ centered on the coordinates ($ x $, $ y $).
If you want to eat the same amount on the $ 2 $ and $ 3 $ days, divide the donut by a straight line passing through the coordinates $ (0,0) $, and the sum of the donut areas in the area on one side of the straight line is on the other side I want it to be equal to the sum of the donut areas in the area of.


Find one slope of such a straight line.

Constraints

The input satisfies the following conditions.

* $ 1 \ lt h \ lt H \ lt 10 ^ 6 $
* $ 1 \ lt w \ lt W \ lt 10 ^ 6 $
* $ 1 \ le y \ le (h / 2 + H-1)-H / 2 $
* $ 1 \ le x \ le (w / 2 + W-1)-W / 2 $
* $ W $, $ H $, $ w $, $ h $ are even numbers

Input

The input is given in the following format.


$ W $ $ H $ $ w $ $ h $ $ x $ $ y $


Output

Output the slope of a straight line in one line. If the absolute error or relative error is $ 10 ^ {-6} $ or less, the answer is correct.

Examples

Input

6000 5000 20 10 400 300


Output

0.75


Input

10 10 8 8 8 8


Output

1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;
using ll = long long;
#define int ll
using PII = pair<int, int>;
template <typename T> using V = vector<T>;
template <typename T> using VV = vector<V<T>>;
template <typename T> using VVV = vector<VV<T>>;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
#define PB push_back

const ll INF = (1LL<<60);
const int MOD = 1000000007;

template <typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template <typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }
template <typename T> bool IN(T a, T b, T x) { return a<=x&&x<b; }
template<typename T> T ceil(T a, T b) { return a/b + !!(a%b); }
template<class S,class T>
ostream &operator <<(ostream& out,const pair<S,T>& a){
  out<<'('<<a.first<<','<<a.second<<')';
  return out;
}
template<class T>
ostream &operator <<(ostream& out,const vector<T>& a){
  out<<'[';
  REP(i, a.size()) {out<<a[i];if(i!=a.size()-1)out<<',';}
  out<<']';
  return out;
}

int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};

signed main(void)
{
  cin.tie(0);
  ios::sync_with_stdio(false);

  int W, H, w, h, x, y;
  cin >> W >> H >> w >> h >> x >> y;

  // 穴の四隅を求める
  int x1 = x - w/2, y1 = y - h/2;
  int x4 = x + w/2, y4 = y + h/2;

  // cout << x1 << " " << y1 << endl << x4 << " " << y4 << endl;

  chmin(x1, W/2);  chmin(x4, W/2); 
  chmax(x1, -W/2); chmax(x4, -W/2);
  chmin(y1, H/2);  chmin(y4, H/2);
  chmax(y1, -H/2); chmax(y4, -H/2);

  // cout << x1 << " " << y1 << endl << x4 << " " << y4 << endl;
  double cx = (x1+x4)/2.0, cy = (y1+y4)/2.0;
  // cout << cx << " " << cy << endl;
  cout << fixed << setprecision(15) << (double)cy/cx << endl;

  return 0;
}