// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

const int N = 1e5 + 5 ;

int p[N],n ;

int rnk[N];

vector < pair < int , pair < int , int > > > v;

int parent(int x){

   if(p[x] == x)

       return x;

   return p[x] = parent(p[x]);

}

bool isConnected(int x , int y){

   return parent(x) == parent(y);

}

void connect(int x , int y){

   if(isConnected(x,y))

       return;

   int pOfx = parent(x);

   int pOfy = parent(y);

   if(rnk[pOfx] > rnk[pOfy]){

       p[pOfy] = pOfx;

   }else if(rnk[pOfy] > rnk[pOfx]){

       p[pOfx] = pOfy;

   }else{

       p[pOfy] = pOfx;

       rnk[pOfx]++;

   }

}

void preDSU(){

   for(int i = 0;i<N;i++){

       p[i] = i;

   }

}



int main(){

    preDSU();

    scanf("%d",&n);

    vector < pair < int ,int > > v1, v2;

    for(int i = 1 ; i <= n ; i++){

        int x,  y;

        scanf("%d%d",&x,&y);

        v1.push_back({x,i});

        v2.push_back({y,i});

    }

    sort(v1.begin(),v1.end());

    sort(v2.begin(),v2.end());

    for(int i = 1 ; i < n ; i++){

        v.push_back({v1[i].first - v1[i - 1].first,{v1[i].second,v1[i - 1].second}});

        v.push_back({v2[i].first - v2[i - 1].first,{v2[i].second,v2[i - 1].second}});

    }

    sort(v.begin(),v.end());

    long long ans = 0 ;

    for(int i = 0 ; i < v.size() ; i++){

        int x = v[i].second.first;

        int y = v[i].second.second;

        long long cost = v[i].first;

        if (isConnected(x, y))

            continue;

        connect(x,y);

        ans += cost;

    }

    cout << ans;

    return 0;

}


// Optimized solution to the same problem (faster):
#include <iostream>

#include <algorithm>

#include <cstdio>

#define rush() int T;cin>>T;while(T--)

#define go(a) while(cin>>a)

#define ms(a,b) memset(a,b,sizeof a)

#define E 1e-8

using namespace std;

typedef long long ll;

const int N=1e5+5;



    int n,m,t;

    int fa[N];

    int i,j,k;

struct point

{

    int x,y,id;

}p[N];

struct Edge

{

    int u,v,valu;

    bool const operator<(const Edge &b)

    const{

        return valu<b.valu;

    }

}edge[N];

int Find(int son)

{

    return son==fa[son] ? son : fa[son]=Find(fa[son]);

}

bool Union(int x,int y)

{

    x=Find(x),y=Find(y);

    if(x==y) return 1;

    else{

        fa[y]=x;

        return 0;

    }

}

bool cmpx(point &a,point &b){

    return a.x<b.x;

}

bool cmpy(point &a,point &b){

    return a.y<b.y;

}

int main()

{

    cin.tie(0);istream::sync_with_stdio(false);

    while(cin>>n){

        for(i=1;i<=n;i++) fa[i]=i;



        for(i=1;i<=n;i++){

            cin>>p[i].x>>p[i].y;

            p[i].id=i;

        }



        int num=0;

        sort(p+1,p+1+n,cmpx);//尽量以x之间的距离为经费建路

        for(i=1;i<n;i++){

            edge[++num].u=p[i].id;

            edge[num].v=p[i+1].id;

            edge[num].valu=min(abs(p[i].x-p[i+1].x),abs(p[i].y-p[i+1].y));

        }

        sort(p+1,p+1+n,cmpy);//尽量以y之间的距离为经费建路

        for(i=1;i<n;i++){

            edge[++num].u=p[i].id;

            edge[num].v=p[i+1].id;

            edge[num].valu=min(abs(p[i].x-p[i+1].x),abs(p[i].y-p[i+1].y));

        }



        stable_sort(edge+1,edge+1+num);

        int ans=0,cnt=0;

        for(i=1;i<=num;i++){

            if(0==Union(edge[i].u,edge[i].v)){//若u，v两点还未连接

                ans+=edge[i].valu;

                cnt++;

            }

            if(cnt==n-1) break;//建路数达到联通所有城市

        }

        cout<<ans<<endl;

    }

}

// Fastest known solution for this problem: