废话不多说，直接上模板

#include "bits/stdc++.h"
#define pb push_back
#define ls l,m,now<<1
#define rs m+1,r,now<<1|1
using namespace std;
typedef long long ll;
typedef pair<int,int> pr;
inline int read() {
    int x=0;
    char c=getchar();
    while(c<'0'||c>'9') c=getchar();
    while(c>='0'&&c<='9') x=(x<<3)+(x<<1)+c-'0', c=getchar();
    return x;
}

const int maxn = 5e3+3;
const int mod = 1e9+7;
const double eps = 1e-9;

struct P{
    int x, y;
}p[maxn], s[maxn];

inline P operator - (P a, P b) {
    return (P){a.x-b.x,a.y-b.y};
}

inline ll X(P a, P b) {
    return a.x*b.y-a.y*b.x;
}

inline double dis(P a, P b) {
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

inline bool operator < (P a, P b) {
    ll t=X(a-p[1],b-p[1]);
    if(t==0) return dis(p[1],a)<dis(p[1],b);
    return t>0;
}

int N, top; //top为凸包顶点数
double ans; // ans为凸包周长

void graham() {
    int t=1;
    for(int i=2; i<=N; ++i)
        if(p[i].y<p[t].y||(p[i].y==p[t].y&&p[i].x<p[t].x)) t=i;
    swap(p[1],p[t]);
    sort(p+2,p+N+1); //将剩下的点按照极角排序！否则就是O(n^2)的算法啦
    s[++top]=p[1];
    s[++top]=p[2]; //关键在于先找到头两个点
    for(int i=3; i<=N; ++i) {
        while(X(s[top]-s[top-1],p[i]-s[top-1])<=0) top--; //常用的一个操作，千万要熟悉！
        s[++top]=p[i];
    }
    s[top+1]=p[1];
    for(int i=1; i<=top; ++i) ans+=dis(s[i],s[i+1]);
}

int main() {
    //ios::sync_with_stdio(false);
    N=read();
    for(int i=1; i<=N; ++i) p[i]=(P){read(),read()}; //从1开始记录
    if(N==1) { //千万记住这两个特判！
        printf("0.00\n"); 
        return 0;
    }
    if(N==2) {
        printf("%.2f\n", dis(p[0],p[1]));
        return 0;
    }
    graham();
    printf("%.2f\n", ans);
}