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

using namespace std;



#define int long long

#define pb push_back

#define fi first

#define se second



const int N=1e5+5;

typedef pair<int,int> ii;

typedef pair<int,ii> iii;



int n,ans;

int pset[N],s[N];

vector<ii> vx,vy;

map<ii,int> ma;

priority_queue<iii,vector<iii>,greater<iii> > pq;



bool how_sort(ii a,ii b){

    if(a.se!=b.se)

            return a.se<b.se;

    return a.fi<b.fi;

}



int unionfind(int u){

    return (u==pset[u]) ? u : (pset[u]=unionfind(pset[u]));

}

bool compare(int u,int v){

    return (unionfind(u)==unionfind(v)) ? true :false;

}

void unionset(int i,int j)

{

    int x=unionfind(i),y=unionfind(j);

    if(s[x]>s[y]){

        pset[y]=x;

        s[x]+=s[y];

    }

    else{

        pset[x]=y;

        s[y]+=s[x];

    }

}

signed main(){

	ios_base::sync_with_stdio(false);

	cin.tie(0);

	int cnt=0;

    cin>>n;

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

        int a,b;

        cin>>a>>b;

        if(ma[ii(a,b)])continue;

        ma[ii(a,b)]=++cnt;

        vx.pb(ii(a,b));

        vy.pb(ii(a,b));

        pset[cnt]=cnt;

        s[cnt]=1;

    }

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

    sort(vy.begin(),vy.end(),how_sort);

    

	// dat09

    for(int i=1;i<vx.size();++i){

        int a=vx[i].fi-vx[i-1].fi;

        pq.push(iii(a,ii(ma[vx[i]],ma[vx[i-1]])));

	

        int b=vy[i].se-vy[i-1].se;

        pq.push(iii(b,ii(ma[vy[i]],ma[vy[i-1]])));

    }

    while(!pq.empty()){

        iii x=pq.top();

        int w=x.fi,u=x.se.fi,v=x.se.se;

        pq.pop();

        if(!compare(u,v)){

            unionset(u,v);

            ans+=w;

        }

    }

    cout<<ans;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>

#include <stdlib.h>

#include <algorithm>



// 全局变量

const int N = 100010 , M = N + N;

int x[N],y[N],id[N],n;



typedef struct ufset *UFset;

struct ufset

{

  int *parent;

  int *root;

};

UFset UFinit(int size)

{

  int e;

  UFset U=(UFset)malloc(sizeof *U);

  U->parent=(int*)malloc((size+1)*sizeof(int));

  U->root=(int*)malloc((size+1)*sizeof(int));

  for(e=1; e<=size; e++)

  {

    U->parent[e]=1;

    U->root[e]=1;

  }

  return U;

}

int UFfind(int e,UFset U)

{

  int i,j = e;

  while(!U->root[j]) j = U->parent[j];

  while(j!=e){

    i = U->parent[e];

    U->parent[e]=j;

    e=i;

  }

  return j;

}

int UFunion(int i,int j,UFset U)

{

  if(U->parent[i]<U->parent[j])

  {

    U->parent[j]+=U->parent[i];

    U->root[i]=0;

    U->parent[i]=j;

    return j;

  }

  else

  {

    U->parent[i]+=U->parent[j];

    U->root[j]=0;

    U->parent[j]=i;

    return i;

  }

}



// 记录优化后的边集数

struct Edge

{

  int u, v, cost;

  bool operator <(const Edge&b) const

  {

    return cost < b.cost;

  }

} edge[M];

int tot;

// 按照X轴排序

bool cmpX(int a, int b)

{

  return x[a] < x[b];

}

// 按照Y轴排序

bool cmpY(int a, int b)

{

  return y[a] < y[b];

}

void add(int c, int a, int b)

{

  edge[tot].cost = c;

  edge[tot].u = a;

  edge[tot++].v = b;

}

void Work()

{

    scanf("%d", &n);

  int i;

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

  {

    scanf("%d%d", x + i, y + i);

    id[i] = i;

  }

  std::sort(id + 1 , id + n + 1, cmpX);

  for (i = 2; i <= n; ++i)

    add(x[id[i]] - x[id[i - 1]], id[i], id[i - 1]);

  std::sort(id + 1 , id + n + 1, cmpY);

  for (i = 2; i <= n; ++i)

    add(y[id[i]] - y[id[i - 1]], id[i], id[i - 1]);

  UFset UFS = UFinit(n);



  // 按照花费排序

  std::sort(edge, edge + tot);

  long long totcost = 0;

  for (i = 0; i < tot; ++i)

  {

    int u = edge[i].u, v = edge[i].v;

    u = UFfind(u,UFS), v = UFfind(v,UFS);

    if (u != v)

    {

      UFunion(u,v,UFS);

      totcost += edge[i].cost;

    }

  }

  printf("%lld\n", totcost);

}

int main()

{

  Work();

  return 0;

}


// Fastest known solution for this problem: