Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7488		Accepted: 2624

Plotters have barberically hammered N nails into an innocent plane shape, so that one can see now only heads. Moreover, pursuing their mean object, they have hammered all the nails into the vertices of a convex polygon. After that they...it is awful... have roped off the nails, so that the shape felt upset (the rope was very thin). They've done it as it is shown in the figure. 
 
Your task is to find out a length of the rope.

There two numbers in the first line of the standard input: N — a number of nails (1 <= N <= 100), and a real number R — a radius of heads of nails. All the heads have the same radius. Further there are N lines, each of them contains a pair of real coordinates (separated by a space) of centers of nails. An absolute value of the coordinates doesn't exceed 100. The nails are described in a clockwise order starting from an arbitrary nail. Heads of different nails don't adjoin.

The standard output should contain in its only line a real number with two digits precision (after a decimal point) — a length of the rope.

4 1

0.0 0.0

2.0 0.0

2.0 2.0

0.0 2.0

14.28

 

 #include <cmath>

 #include <cstdio>

 #include <iostream>

 #include <algorithm>

 #define INF 0x3f3f3f3f

 #define pi 3.1415

 using namespace std;

 const int MAXN = ;

 double x[MAXN], y[MAXN];

 double ans, R;

 int N;

 double s(double x1, double x2, double y1, double y2)

 {

     return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));

 }

 int main()

 {

     scanf("%d%lf", &N, &R);

     for(int i = ; i <= N; i++)

     {

         scanf("%lf %lf", &x[i], &y[i]);

     }

     for(int i = ; i <= N; i++)

     {

         ans+=s(x[i], x[i-], y[i], y[i-]);

     }

     ans+=s(x[], x[N], y[], y[N]);

     ans+=(*pi*R);

     printf("%.2lf\n", ans);

     return ;

 }