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

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>



using namespace std;

using namespace __gnu_pbds;



#define fi first

#define se second

#define mp make_pair

#define pb push_back

#define fbo find_by_order

#define ook order_of_key



typedef long long ll;

typedef pair<ll,ll> ii;

typedef vector<int> vi;

typedef long double ld;

typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;

bool marked[100001];

vector<ii> v[100001];  //adjList



ll prim(int k){

    priority_queue<ii, vector<ii>, greater<ii> > q;

    ll u;

    long long minimumCost = 0;

    ii p;

    q.push(mp(0, k));

    while(!q.empty())

    {

        p = q.top();

        q.pop();

        k = p.second;

        if(marked[k] == true)

            continue;

        minimumCost += p.fi;

        marked[k] = true;

        for(ll i = 0;i < v[k].size();++i)

        {

            u = v[k][i].se;

            if(marked[u] == false)

                q.push(v[k][i]);

        }

    }

    return minimumCost;

}



int main() {

    std::ios::sync_with_stdio(false);

    cin.tie(NULL);

    long long n;

    cin>>n;

    vector<ii> x;

    vector<ii> y;

    for(ll i =0;i<n;i++){

        ll a,b;

        cin>>a>>b;

        x.pb(mp(a,i));

        y.pb(mp(b,i));

    }

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

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

    for(ll i=0;i<n-1;i++){

        v[x[i].se].pb(mp(x[i+1].fi-x[i].fi,x[i+1].se));

        v[x[i+1].se].pb(mp(x[i+1].fi-x[i].fi,x[i].se));

        v[y[i].se].pb(mp(y[i+1].fi-y[i].fi,y[i+1].se));

        v[y[i+1].se].pb(mp(y[i+1].fi-y[i].fi,y[i].se));

    }

    ll ans=prim(0);

    cout<<ans;

    return 0;

}


// 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: