Codeforces Round #287 (Div. 2)B. Amr and Pins

B. Amr and Pins

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Amr loves Geometry. One day he came up with a very interesting problem.

Amr has a circle of radius r and center in point (x, y). He wants the circle center to be in new position (x', y').

In one step Amr can put a pin to the border of the circle in a certain point, then rotate the circle around that pin by any angle and finally remove the pin.

Help Amr to achieve his goal in minimum number of steps.

Input

Input consists of 5 space-separated integers r, x, y, x' y' (1 ≤ r ≤ 105,  - 105 ≤ x, y, x', y' ≤ 105), circle radius, coordinates of original center of the circle and coordinates of destination center of the circle respectively.

Output

Output a single integer — minimum number of steps required to move the center of the circle to the destination point.

Examples

input

Copy

2 0 0 0 4

output

Copy

1

input

Copy

1 1 1 4 4

output

Copy

3

input

Copy

4 5 6 5 6

output

Copy

0

Note

In the first sample test the optimal way is to put a pin at point (0, 2) and rotate the circle by 180 degrees counter-clockwise (or clockwise, no matter).

题目给定一个起始圆心和圆的半径以及终点圆心，可以在圆内选定任意一点进行任意角度的旋转。

直接计算一下起始圆心到终点圆心的距离，然后用距离去除圆的半径，得到的结果向上取整即可。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
	double r,x,y,xx,yy;
	while(cin>>r>>x>>y>>xx>>yy)
	{
		double res=sqrt((xx-x)*(xx-x)+(yy-y)*(yy-y));
			res/=r*2;
			int ans=(int)res;
			if(ans!=res)
				ans++;
			cout<<ans<<endl;
	}
	return 0;
}