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)

//This is getting accepted!
// I HATE BUG
// God Of The Bugs
// 12/11/2016
#include<bits/stdc++.h>

using namespace std;

#define ms(s, n) memset(s, n, sizeof(s))
#define FI first
#define SE second
#define pb push_back
#define mp make_pair
#define ll long long
#define sz(a) ((int)(a).size())
#define __builtin_popcount __builtin_popcounll
#define ld long double

typedef pair<int, int> pii;
typedef pair<double, double> pdd;
typedef pair<int, pii> ppi;

const double PI = acos(0) * 2;
const double EPS = 1e-8;
const ll MOD = 1e9 + 7;
const int MAXN = 1e5 + 5;
const int oo = 1e9;
const double foo = 1e30;

template<class T> int getbit(T s, int i) { return (s >> i) & 1; }
template<class T> T onbit(T s, int i) { return s | (T(1) << i); }
template<class T> T offbit(T s, int i) { return s & (~(T(1) << i)); }
template<class T> int cntbit(T s) { return __builtin_popcounll(s);}
template<class T> T sqr(T x) { return x * x; }

inline void addmod(int& a, int val, int p = MOD) {if ((a = (a + val)) >= p) a -= p;}
inline void submod(int& a, int val, int p = MOD) {if ((a = (a - val)) < 0) a += p;}
inline int mult(int a, int b, int p = MOD) {return (ll) a * b % p;}

int par[MAXN], _size[MAXN];
vector<pii> X, Y;
vector<pair<int, pii> > ed;
int n, x[MAXN], y[MAXN];

int getpar(int u) {
	while (u != par[u]) {
		par[u] = par[par[u]];
		u = par[u];
	}
	return u;
}

int chich(int a, int b) {
	int A = getpar(a), B = getpar(b);
	if (A == B) return 0;
	if (_size[A] > _size[B]) swap(A, B);
	par[A] = B;
	_size[B] += _size[A];
	return 1;
}

int main() {
//#ifndef ONLINE_JUDGE
//    freopen("inp.txt", "r", stdin);
////    freopen("out.txt", "w", stdout);
//#endif

	cin >> n;
	
	for (int i=0; i<n; i++) cin >> x[i] >> y[i];
	
	for (int i=0; i<n; i++) {
		X.pb(mp(x[i], i));
		Y.pb(mp(y[i], i));
	}
	
	sort(X.begin(), X.end());
	sort(Y.begin(), Y.end());
	
	for (int i=1; i<n; i++) {
		ed.pb(mp(X[i].FI - X[i - 1].FI, mp(X[i].SE, X[i - 1].SE)));
		ed.pb(mp(Y[i].FI - Y[i - 1].FI, mp(Y[i].SE, Y[i - 1].SE)));
	}

	for (int i=0; i<n; i++) {
		par[i] = i;
		_size[i] = 1;
	}
	sort(ed.begin(), ed.end());
	
	ll ans = 0;
	for (int i=0; i<sz(ed); i++) {
		if (chich(ed[i].SE.FI, ed[i].SE.SE)) ans += (ll) ed[i].FI;
	}
	
	cout << ans;
}