/*
    ans[sta][i]点的集合为sta，且最后一个点是i的最短路 
*/
#include<cmath>
#include<cstdio>
#include<iostream>
using namespace std;
int n,num1[1<<20];
double x[21],y[21],dis[21][21],ans[1<<20][21];
int main(){
    //freopen("Cola.txt","r",stdin);
    freopen("linec.in","r",stdin);freopen("linec.out","w",stdout);
    scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%lf%lf",&x[i],&y[i]);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++){
            if(i==j)continue;
            dis[i][j]=sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
        }
    for(int i=0;i<=n;i++)num1[1<<i]=i+1;
    for(int i=1;i<(1<<n);i++){
        if(!num1[i]){
            for(int j=i;j;j-=j&-j){
                ans[i][num1[j&-j]]=0x3f3f3f3f;
                for(int k=i-(j&-j);k;k-=(k&-k))
                    if (ans[i][num1[j&-j]]>ans[i-(j&-j)][num1[k&(-k)]]+dis[num1[j&-j]][num1[k&-k]])
                        ans[i][num1[j&-j]]=ans[i-(j&-j)][num1[k&(-k)]]+dis[num1[j&-j]][num1[k&-k]];
            }
        }
    }
    double Ans=0x7fffffff;
    for(int i=1;i<=n;i++)Ans=min(Ans,ans[(1<<n)-1][i]);
    printf("%.2lf",Ans);
}