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

using namespace std;

const long long llINF = 9223372036854775807;

const int INF = 2147483647;

const int maxn = 1e5 + 7;

const int maxm = 1e6 + 7;

const int mod = 1e9 + 7;

const double pi = acos(-1.0);

const double eps = 1e-10;

int n;

int fa[maxn];

struct node

{

    long long u, v, w;

};

vector<node> edge;

bool operator<(node a, node b)

{

    return a.w < b.w;

}

struct node2

{

    long long x, y, id;

} city[maxn];



bool cmp1(node2 a, node2 b)

{

    return a.x < b.x;

}



bool cmp2(node2 a, node2 b)

{

    return a.y < b.y;

}

void init()

{

    for (int i = 1; i < maxn; i++)

        fa[i] = i;

}

int get(int x)

{

    return fa[x] == x ? x : fa[x] = get(fa[x]);

}

long long ans;

int main(int argc, char const *argv[])

{

    init();

    scanf("%d", &n);

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

    {

        scanf("%lld%lld", &city[i].x, &city[i].y);

        city[i].id = i;

    }

    sort(city + 1, city + 1 + n, cmp1);

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

        edge.push_back(node{city[i].id, city[i + 1].id, abs(city[i].x - city[i + 1].x)});



    sort(city + 1, city + 1 + n, cmp2);

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

        edge.push_back(node{city[i].id, city[i + 1].id, abs(city[i].y - city[i + 1].y)});



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



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

    {

        int x=get(edge[i].u);

        int y=get(edge[i].v);

        if(x==y)

            continue;

        fa[x]=y;

        ans+=edge[i].w;

    }

    

    printf("%lld\n",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;

}

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: