Problem:
Score : 500 points

Problem Statement

There are N towns on a plane. The i-th town is located at the coordinates (x_i,y_i). There may be more than one town at the same coordinates.

You can build a road between two towns at coordinates (a,b) and (c,d) for a cost of min(|a-c|,|b-d|) yen (the currency of Japan). It is not possible to build other types of roads.

Your objective is to build roads so that it will be possible to travel between every pair of towns by traversing roads. At least how much money is necessary to achieve this?

Constraints

2 ≤ N ≤ 10^5

0 ≤ x_i,y_i ≤ 10^9

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
x_1 y_1
x_2 y_2
:
x_N y_N

Output

Print the minimum necessary amount of money in order to build roads so that it will be possible to travel between every pair of towns by traversing roads.

Sample Input 1

3
1 5
3 9
7 8

Sample Output 1

3

Build a road between Towns 1 and 2, and another between Towns 2 and 3. The total cost is 2+1=3 yen.

Sample Input 2

6
8 3
4 9
12 19
18 1
13 5
7 6

Sample Output 2

8

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<list>
#include<cstdint>
#include<bits/stdc++.h>
#include<queue>
#include<set>

#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define rep_r(i, n) for(int i = n-1; i >= 0; i--)
#define rep1(i, n) for(int i = 1; i <= (int)(n); i++)
#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((ll)(x).size())
#define bit(n) (1LL<<(n))
#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() );

#define INF bit(60)
#define pb push_back
#define mod 1000000007

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

using namespace std;
using uif=uint_fast64_t;
using ll=long long int;

int qp(int a,ll b){int ans=1;do{if(b&1)ans=1ll*ans*a%mod;a=1ll*a*a%mod;}while(b>>=1);return ans;}
int qp(int a,ll b,int mo){int ans=1;do{if(b&1)ans=1ll*ans*a%mo;a=1ll*a*a%mo;}while(b>>=1);return ans;}

int gcd(int a,int b){return b?gcd(b,a%b):a;}
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};


int main(void){

  ll n;
  cin>>n;
  vector<ll> x(n);
  vector<ll> y(n);
  vector<ll> mincost(n);
  vector<bool> used(n);

  rep(i,n){
    cin>>x[i];
    cin>>y[i];
    mincost[i] = INF;
    used[i] = false;
  }
  mincost[0]=0;
  ll ans=0;

  while(true){
    ll v=-1;
    rep(u,n){
      if(!used[u] && (v==-1 || mincost[u] < mincost[v])){
        v=u;
        //cout << "v update into " << v << endl;
      }
    }
    if(v == -1){
      //cout << "break!" << endl;
      break;
    }
    used[v]=true;
    //cout << v << ":" << mincost[v] << endl;
    ans+=mincost[v];

    rep(u,n){
      if(u==v) continue;
      mincost[u] = min(mincost[u], min(abs(x[u]-x[v]),abs(y[u]-y[v])) );
    }

  }

  cout << ans << endl;

  return 0;
}