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;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;
typedef pair<ll, pair<ll, ll> > P3;

const ll MOD = ll(1e9 + 7);
const ll LLINF = LLONG_MAX;
const int IINF = INT_MAX;
const int MAX_N = int(1e5) + 5;
const double EPS = 1e-8;
const int di[] = {0, 1, 0, -1}, dj[] = {1, 0, -1, 0};
string dir[] = {"R", "D", "L", "U"};
#define REP(i, n) for (int i = 0; i < n; i++)
#define REPR(i, n) for (int i = n; i >= 0; i--)
#define FOR(i, m, n) for (ll i = m; i < n; i++)
#define ALL(v) (v).begin(), (v).end()

int main() {
    double W, H, w, h, x, y;
    cin >> W >> H >> w >> h >> x >> y;
    double xh,xl,yh,yl;
    xh = min(W/2,x+w/2);
    xl = max(-W/2,x-w/2);
    yh = min(H/2,y+h/2);
    yl = max(-H/2,y-h/2);
    printf("%.8lf\n",(yh+yl)/(xh+xl));
    return 0;
}