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 ld = long double;

int main()
{
	ios::sync_with_stdio(false), cin.tie(0);
	ld W, H, w, h, x, y;
	cin >> W >> H >> w >> h >> x >> y;
	ld tx[] = { x - w / 2, x + w / 2, x - w / 2, x + w / 2 };
	ld ty[] = { y - h / 2, y - h / 2, y + h / 2, y + h / 2 };
	for (int i = 0; i < 4; i++) {
		tx[i] = min(tx[i], W / 2);
		tx[i] = max(tx[i], -W / 2);
		ty[i] = min(ty[i], H / 2);
		ty[i] = max(ty[i], -H / 2);
	}
	ld gx = 0, gy = 0;
	for (int i = 0; i < 4; i++) {
		gx += tx[i];
		gy += ty[i];
	}
	cout << fixed << setprecision(10) << (ld)gy / gx << endl;
	return 0;
}