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: Accepted)

#include <iostream>
#include <queue>
#include <assert.h>   
#include <vector>
#include <algorithm>
#include <time.h>
#include <numeric>
#include <string>
#include <cstring>
#include <stack>
#include <unordered_set>
#include <unordered_map>
#include <set>
#include <map>
#include <numeric>
#include <deque>
#include <iomanip>
#include <cstdlib>
#include <climits>
#include <cmath>
#include <cstdio>
using namespace std;


//#define double long double
#define int long long
//#define int unsigned long long

#define ll long long
#define ld long double 
#define pi pair<int, int> 
#define pl pair<ll,ll>  
#define pd pair<ld,ld> 
#define str string  
#define mp make_pair
#define vi vector<int> 
#define vl vector<ll> 
#define vd vector<ld> 
#define vs vector<str> 
#define vpi vector<pi> 
#define vpl vector<pl> 
#define vpd vector<pd> 
#define si set<int> 
#define FF(i,j,n) for(signed i=j;i<=n;i++)  
#define DD(i,j,n) for(signed i=j;i>=n;i--) 
#define F(i,j,n) for(signed i=j;i<n;i++)   
#define D(i,j,n) for(signed i=j;i>n;i--) 
#define mem(a,b) memset(a,b,sizeof(a))
#define ndl cout << endl
#define ull unsigned long long 

const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3fLL;


#define sz(x) (signed)x.size()
#define all(x) begin(x), end(x)
#define rall(x) (x).rbegin(), (x).rend() 
#define rsz resize
#define ins insert 
#define ft front() 
#define bk back() 
#define fi first
#define se second
#define pf push_front 
#define pb push_back
#define eb emplace_back 
#define lb lower_bound 
#define ub upper_bound 
#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);

//math


const int mod = 1e9 + 7;

ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
int pct(int x) { return __builtin_popcount(x); } 
int bit(int x) { return 31-__builtin_clz(x); } // floor(log2(x)) 
int cdiv(int a, int b) { return a/b+!(a<0||a%b == 0); } // division of a by b rounded up, assumes b > 0 


//DEBUG, compile with -DLOCAL
void DBG() { cerr << "]" << endl; }
template<class H, class... T> void DBG(H h, T... t) {
  cerr << to_string(h); if (sizeof...(t)) cerr << ", ";
  DBG(t...); }
#ifdef LOCAL 
#define dbg(...) cerr << "[" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__)
#define HERE cout << "===GOTHERE===" << endl
#else
#define dbg(...) 42
#define HERE 42
#endif


 /* ============== END OF HEADER ============== */





const int maxn = 1e5*3+20;

int f[maxn];

int n,_,q,k;

struct P{
	int x; int y; int id;
};
vector<P> pts;
vector<P> byx;
vector<P> byy;

struct Edge{
	int fr; int to; int w;
};
vector<Edge> edges;


int dist(P p1, P p2){
	return min(abs(p1.x - p2.x), abs(p1.y - p2.y));
}

int find(int x){
    if(f[x]==x)return x;
    return f[x]=find(f[x]);
}
void unite(int x,int y){
    x=find(x);y=find(y);
    if(x==y)return;
    f[x]=y;
}

void solve(){
	cin >> n;
	FF(i,1,n){
		int t1,t2; cin >> t1 >> t2;
		pts.eb((P){t1,t2,i});
	}
	byx = pts;
	byy = pts;
	sort(all(byx), [](P p1, P p2){
		return p1.x < p2.x;
	});
	sort(all(byy), [](P p1, P p2){
		return p1.y < p2.y;
	});

	//build edges
	F(i,0,sz(byx)){
		P curr = byx[i];
		if(i-1 >= 0){
			P lft = byx[i-1];
			Edge ne = (Edge){curr.id, lft.id, dist(lft,curr)};
			edges.eb(ne);
		}
		if(i+1 < sz(byx)){
			P rt = byx[i+1];
			Edge ne = (Edge){curr.id, rt.id, dist(rt,curr)};
			edges.eb(ne);
		}
	}
	F(i,0,sz(byy)){
		P curr = byy[i];
		if(i-1 >= 0){
			P lft = byy[i-1];
			Edge ne = (Edge){curr.id, lft.id, dist(lft,curr)};
			edges.eb(ne);
		}
		if(i+1 < sz(byy)){
			P rt = byy[i+1];
			Edge ne = (Edge){curr.id, rt.id, dist(rt,curr)};
			edges.eb(ne);
		}
	}

	for(int i = 0; i < n; i++){
		f[i] = i;
	}
	sort(all(edges), [](Edge e1, Edge e2){
		return e1.w < e2.w;
	});	
	int sum = 0;
	dbg(sz(edges));
	for(Edge te : edges){
		if(find(te.fr) != find(te.to)){
			f[find(te.fr)] = f[find(te.to)];
			sum += te.w;
		}
	}
	cout << sum;


}

signed main(){
  //clock_t t1 = clock();  
	IOS
	solve();
  
  //cout << "THE ALGORITHM TOOK: " << (clock() - t1) * 1.0 / CLOCKS_PER_SEC * 1000 << "ms"<< endl;
	return 0;
}