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

using namespace std;

#define pb push_back

#define mp make_pair

#define fi first

#define se second

int n,pset[100007];

long long x[100007],y[100007],ans=0;

pair < pair < long long,long long > , int > p[100007];

vector < pair < long long, pair < int,int > > > cur;

int fset(int x){

    if(pset[x]==x) return x;

    return pset[x]=fset(pset[x]);

}

int main(){

    ios_base::sync_with_stdio(0);

    cin>>n;

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

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

        p[i]=mp(mp(x[i],y[i]),i);

        pset[i]=i;

    }

    sort(p+1,p+1+n);

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

        cur.pb(mp(p[i].fi.fi-p[i-1].fi.fi,mp(p[i].se,p[i-1].se)));

    }

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

        p[i]=mp(mp(y[i],x[i]),i);

    }

    sort(p+1,p+1+n);

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

        cur.pb(mp(p[i].fi.fi-p[i-1].fi.fi,mp(p[i].se,p[i-1].se)));

    }

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

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

        int a=cur[i].se.fi,b=cur[i].se.se;

        long long val=cur[i].fi;

        a=fset(a);

        b=fset(b);

        if(a!=b){

            pset[b]=a;

            ans+=val;

        }

    }

    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;

}

int main()

{

  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);

  return 0;

}

// Fastest known solution for this problem: