题干：

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

2 0 0 0 4

Output

1

Input

1 1 1 4 4

Output

3

Input

4 5 6 5 6

Output

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).

解题报告：

     不难发现，我们走一步可以走的最大距离就是走一个直径，不难证明，小于一个直径的距离，我们都可以通过调整旋转点来一步到位，所以说啊，，直接贪心就出来了。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;

int main()
{
	ll r,x,xx,y,yy;
	cin>>r>>x>>y>>xx>>yy;
	r*=2;
	double dis = sqrt((xx-x)*(xx-x) + (yy-y)*(yy-y));
	if(x == xx && y == yy) puts("0");
	else {
		printf("%lld\n",(ll)ceil(dis/r));
	} 


	return 0 ;
 }