// Original C++ solution (slower):
#include<iostream>

#include<string>

#include<algorithm>

#include <string.h>

#include <stdio.h>

#include <math.h>

#include <set>

#include <queue>

#include <stack>

#include <map>



using namespace std;

typedef int status;

typedef long long LL;

const int INF = int(1e9);

const int MAX = 100005;



struct S

{

    int x,wei;

}A[MAX],B[MAX];



struct Bian

{

    int x,y;

    int len;

}biao[MAX*10];

int num = 0;



bool cmp(S a,S b)

{

    return a.x<b.x;

}



bool cmp1(Bian a,Bian b)

{

    return a.len<b.len;

}



int N;



int visit[MAX] = {0};



int Find(int x)

{

    if(visit[x]==x)

        return x;

    else

        return visit[x] = Find(visit[x]);

}



int mix(int x,int y)

{

    int Tx = Find(x);

    int Ty = Find(y);

    if(Tx!=Ty)

    {

        visit[Tx] = Ty;

        return 1;

    }

    return 0;



}





int main()

{

    cin>>N;

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

    {

        A[i].wei = B[i].wei = i;

        scanf("%d %d",&A[i].x,&B[i].x);

    }

    sort(A+1,A+1+N,cmp);sort(B+1,B+1+N,cmp);

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

    {

        biao[num].x = A[i].wei;

        biao[num].y = A[i+1].wei;

        biao[num++].len = A[i+1].x - A[i].x;

        biao[num].x = B[i].wei;

        biao[num].y = B[i+1].wei;

        biao[num++].len = B[i+1].x - B[i].x;

    }

    sort(biao,biao+num,cmp1);



    LL sum = 0;

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

        visit[i] = i;

    for(int i = 0;i<num;i++)

    {

        Bian temp = biao[i];

        if(mix(temp.x,temp.y)==1)

        {

            sum += temp.len;

        }

    }



    cout<<sum<<endl;



    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: