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

using namespace std;



int p[100001];



struct node{

	int x, y, id;

}nodes[100001];



struct edge{

	int u, v, cost;

}mst[100001];



int find(int u){

	if(p[u] == u)

		return p[u];

	else 

		return p[u] = find(p[u]);

}



bool cmp1(node &a, node &b){

	return a.x < b.x;

}



bool cmp2(node &a, node &b){

	return a.y < b.y;

}



bool cmp3(edge &a, edge &b){

	return a.cost < b.cost;

}



signed main(){

	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

	

	int n, k = 0;

	cin >> n;

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

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

		nodes[i].id = i;

	}

	

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

	

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

		mst[k].u = nodes[i].id;

		mst[k].v = nodes[i+1].id;

		mst[k].cost = abs(nodes[i].x - nodes[i+1].x);

		k++;

	}

	

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

	

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

		mst[k].u = nodes[i].id;

		mst[k].v = nodes[i+1].id;

		mst[k].cost = abs(nodes[i].y - nodes[i+1].y);

		k++;

	}

	

	sort(mst, mst + k, cmp3);

	

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

		p[i]=i;

		

	long long ans=0;

	for(int i = 0; i < k; i++) {

		int a = mst[i].u, b = mst[i].v;

		a = find(a), b = find(b);

		if(a != b) {

			ans += mst[i].cost;

			p[a] = b;

		}

	}

	

	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: