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 <iostream>
#include <iomanip>
using namespace std;

int main(){
    double bw,bh,sw,sh,x,y;
    cin >> bw >> bh >> sw >> sh >> x >> y;
    double midx = (x -sw/2 +min(x +sw/2, bw/2) ) /2;
    double midy = (y -sh/2 +min(y +sh/2, bh/2) ) /2;
    cout << fixed << setprecision(10);
    cout << midy /midx << endl;
    return 0;
}