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 <bits/stdc++.h>
#define LOOP(n) for (int _i = 0; _i < (n); _i++)
#define REP(i, n) for (int i = 0; i < (n); ++i)
#define RREP(i, n) for (int i = (n); i >= 0; --i)
#define FOR(i, r, n) for (int i = (r); i < (n); ++i)
#define ALL(obj) begin(obj), end(obj)
using namespace std;
using ll = long long;
using ull = unsigned long long;
using P = pair<ll, ll>;
// 多次元 vector 生成
template <class T>
vector<T> make_vec(size_t a) {
    return vector<T>(a, 0);
}
template <class T, class... Ts>
auto make_vec(size_t a, Ts... ts) {
    return vector<decltype(make_vec<T>(ts...))>(a, make_vec<T>(ts...));
}
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;
}

/* UnionFind：素集合系管理の構造体(union by rank)
    isSame(x, y): x と y が同じ集合にいるか。 計算量はならし O(α(n))
    unite(x, y): x と y を同じ集合にする。計算量はならし O(α(n))
*/
struct UnionFind {  // The range of node number is u 0 v n-1
    vector<int> rank, parents;
    UnionFind() {}
    UnionFind(int n) {  // make n trees.
        rank.resize(n, 0);
        parents.resize(n, 0);
        for (int i = 0; i < n; i++) {
            makeTree(i);
        }
    }
    void makeTree(int x) {
        parents[x] = x;  // the parent of x is x
        rank[x] = 0;
    }
    bool isSame(int x, int y) { return findRoot(x) == findRoot(y); }
    void unite(int x, int y) {
        x = findRoot(x);
        y = findRoot(y);
        if (rank[x] > rank[y]) {
            parents[y] = x;
        } else {
            parents[x] = y;
            if (rank[x] == rank[y]) {
                rank[y]++;
            }
        }
    }
    int findRoot(int x) {
        if (x != parents[x]) parents[x] = findRoot(parents[x]);
        return parents[x];
    }
};

// 辺の定義
struct Edge {
    long long u;
    long long v;
    long long cost;
};
bool comp_e(const Edge &e1, const Edge &e2) { return e1.cost < e2.cost; }

/* Kruskal :クラスカル法で minimum spanning tree を求める構造体
    入力: 辺のvector, 頂点数V
    最小全域木の重みの総和: sum
    計算量: O(|E|log|V|)
*/
struct Kruskal {
    UnionFind uft;
    long long sum;  // 最小全域木の重みの総和
    vector<Edge> edges;
    int V;
    Kruskal(const vector<Edge> &edges_, int V_) : edges(edges_), V(V_) { init(); }
    void init() {
        sort(edges.begin(), edges.end(), comp_e);
        uft = UnionFind(V);
        sum = 0;
        for (auto e : edges) {
            if (!uft.isSame(e.u, e.v)) {
                uft.unite(e.u, e.v);
                sum += e.cost;
            }
        }
    }
};

int main() {
    int N;
    cin >> N;
    vector<tuple<int, int, int>> px(N), py(N);
    REP(i, N) {
        ll x, y;
        cin >> x >> y;
        px.at(i) = make_tuple(x, y, i);
        py.at(i) = make_tuple(y, x, i);
    }
    vector<Edge> edges;
    sort(ALL(px));
    for (int i = 0; i < N - 1; i++) {
        edges.push_back({get<2>(px[i]), get<2>(px[i + 1]), get<0>(px[i + 1]) - get<0>(px[i])});
    }
    sort(ALL(py));
    for (int i = 0; i < N - 1; i++) {
        edges.push_back({get<2>(py[i]), get<2>(py[i + 1]), get<0>(py[i + 1]) - get<0>(py[i])});
    }
    Kruskal krs(edges, N);
    cout << krs.sum << endl;
    return 0;
}