/*

需要好大的空间.....

而且lowbit理解的不是很好

先放到博客里 以后慢慢研究

*/

#include<iostream>

#include<cstdio>

#include<cstring>

#include<cmath>

#define inf 999999999;

#define maxn 1050000

using namespace std;

int n,a[maxn];

double x[],y[],g[][],f[maxn][],ans=inf;

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

{

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

}

int l(int x)

{

    return x&(-x);

}

int main()

{

    freopen("linec.in","r",stdin);

    freopen("linec.out","w",stdout);

    scanf("%d",&n);

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

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

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

      for(int j=;j<=n;j++)

        g[i][j]=Dis(x[i],y[i],x[j],y[j]);

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

      a[<<i]=i+;

    for(int i=;i<(<<n);i++)if(!a[i])

      for(int j=i;j;j-=l(j))//枚举状态i的每一个1

        {

          f[i][a[l(j)]]=inf;//更新此状态

          for(int k=i-l(j);k;k-=l(k))

            f[i][a[l(j)]]=min(f[i][a[l(j)]],f[i-l(j)][a[l(k)]]+g[a[l(j)]][a[l(k)]]);

        }

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

      ans=min(ans,f[(<<n)-][i]);

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

    return ;

}