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

#include <algorithm>

#include <cmath>

#include <cstring>

#include <cstdio>

#include <map>

#include <stack>

#include <queue>

#include <set>

#include <cstdlib>

#include <stack>

#include <vector>

#define INF 0x3f3f3f3f

#define MOD 1000000009

# define FOR(i,a,n) for(register int i=a; i<=n; ++i)

# define FDR(i,a,n) for(register int i=a; i>=n; --i)

typedef long long ll;



using namespace std;

struct A

{

    int l;

    int r;

    int q;

};

struct B

{

    int a;

    int b;

    int num;

};

B bb[300005];

A aa[300005];

int a[100005], b[100005];

int f[100005];

int finds(int x)

{

    return f[x] == x?x:f[x] = finds(f[x]);

}

int cmp1(B a, B b)

{

    return a.a < b.a;

}

int cmp2(B a, B b)

{

    return a.b < b.b;

}

int cmp3(A a, A b)

{

    return a.q < b.q;

}

int main()

{

    int n;

    scanf("%d", &n);

    FOR(i,1,n)

    {

        scanf("%d%d", &bb[i].a, &bb[i].b);

        bb[i].num = i;

    }

    int ji = 0;

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

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

    {

        aa[ji].l = bb[i].num;

        aa[ji].r = bb[i+1].num;

        aa[ji++].q = min(abs(bb[i].a-bb[i+1].a), abs(bb[i].b-bb[i+1].b));

    }

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

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

    {

        aa[ji].l = bb[i].num;

        aa[ji].r = bb[i+1].num;

        aa[ji++].q = min(abs(bb[i].a-bb[i+1].a), abs(bb[i].b-bb[i+1].b));

    }

    sort(aa, aa+ji, cmp3);

    FOR(i,1,n)

    f[i] = i;

    ll ans = 0;

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

    {

        int x = finds(aa[i].l);

        int y = finds(aa[i].r);

        if(x != y)

        {

            f[x] = y;

            ans += aa[i].q;

        }

    }

    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;

}

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: