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 pair<int, int> P;

int MOD = 1e9 + 7;
ll INF = 1e18;

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

int main(){
    int W, H, w, h; cin>>W>>H>>w>>h;
    double x, y; cin>>x>>y;
    
    double X=x, Y=y;
    if(x+w/2>W/2 && y+h/2>H/2){
        X=((x-w/2)+W/2)/2;
        Y=((y-h/2)+H/2)/2;
    }
    else if(y+h/2>=H/2){
        Y=((y-h/2)+H/2)/2;
    }
    else if(x+w/2>=W/2){
        X=((x-w/2)+W/2)/2;
    }
    cout << fixed << setprecision(10) << Y/X << endl;
}