1 //2016.10.2
 2 #include <iostream>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <cmath>
 6 #include <algorithm>
 7 #define N 105
 8 #define eps 1e-8
 9 
10 using namespace std;
11 
12 struct point
13 {
14     double x, y;
15     point(){}
16     point(double a, double b):x(a), y(b){}
17     point operator-(point a){//向量减法
18         return point(x-a.x, y-a.y);
19     }
20     point operator+(point a){//向量加法
21         return point(x+a.x, y+a.y);
22     }
23     double operator*(point a){//向量叉积
24         return x*a.y-y*a.x;
25     }
26     bool operator<(const point a)const{
27         if(fabs(x-a.x)<eps)return y<a.y;//浮点数的判等不能直接用‘==’直接比较
28         return x<a.x;
29     }
30     double len(){//向量的模
31         return sqrt(x*x+y*y);
32     }
33 }p[N], s[N];//p为点，s为栈
34 
35 double cp(point a, point b, point o)//向量oa，ob叉积
36 {
37     return (a-o)*(b-o);
38 }
39 
40 void Convex(point *p, int &n)//Graham扫描法，栈内为所有凸包点
41 {
42     sort(p, p+n);
43     int top, m;
44     s[0] = p[0]; s[1] = p[1]; top = 1;
45     for(int i = 2; i < n; i++)//从前往后扫
46     {
47         while(top>0 && cp(p[i], s[top], s[top-1])>=0)top--;
48         s[++top] = p[i];
49     }
50     m = top;
51     s[++top] = p[n-2];
52     for(int i = n-3; i >= 0; i--)//从后往前扫
53     {
54         while(top>m && cp(p[i], s[top], s[top-1])>=0)top--;
55         s[++top] = p[i];
56     }
57     n = top;
58 }
59 
60 int main()
61 {
62     int n;
63     while(scanf("%d", &n)!=EOF && n)
64     {
65         for(int i = 0; i < n; i++)
66               scanf("%lf%lf", &p[i].x, &p[i].y);
67         sort(p, p+n);
68         int cnt = 0;
69         for(int i = 1; i < n; i++)//去掉重复的点
70               if(fabs(p[i].x-p[cnt].x)>eps || fabs(p[i].y-p[cnt].y)>eps)
71                   p[++cnt] = p[i];
72         cnt++;
73         if(cnt == 1){
74             printf("0.00\n");continue;
75         }else if(cnt==2){
76             printf("%.2lf\n", (p[1]-p[0]).len());continue;
77         }
78         Convex(p, cnt);
79         double ans = 0;
80         s[cnt] = s[0];
81         for(int i = 0; i < cnt; i++)ans+=(s[i+1]-s[i]).len();
82         printf("%.2lf\n", ans);
83     }
84 
85     return 0;
86 }