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 REP(i, x, n) for (int i = x; i < (int)(n); i++)
#define rep(i, n) REP(i, 0, n)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) (int)(x.size())
#define popcount(x) __builtin_popcount(x)
#define popcountll(x) __builtin_popcountll(x)
#define uniq(x) x.erase(unique(x.begin(), x.end()), x.end())
#define F first
#define S second
#define mp make_pair
#define eb emplace_back
#define pii pair<int, int>
#define pll pair<ll, ll>

using namespace std;

typedef long long ll;

/* --- INFの値が適切か確認する --- */
// const int INF = 1 << 30;
// const ll INF = 1LL << 60;
// const int MOD = 1000000007;

class UnionFindTree {
 private:
  int nodeSize;
  vector<int> parent;
  vector<int> rank;
  vector<int> treeSize;

 public:
  UnionFindTree(int ns) {
    nodeSize = ns;
    parent = vector<int>(nodeSize);
    rank = vector<int>(nodeSize, 0);
    treeSize = vector<int>(nodeSize, 1);

    for (int i = 0; i < nodeSize; i++) {
      parent[i] = i;
    }
  }

  int find(int x) {
    if (parent[x] == x) {
      return x;
    } else {
      return parent[x] = find(parent[x]);
    }
  }

  bool unite(int x, int y) {
    x = find(x);
    y = find(y);

    if (x == y) {
      return false;
    }

    if (rank[x] < rank[y]) {
      parent[x] = y;
      treeSize[y] += treeSize[x];
    } else {
      parent[y] = x;
      treeSize[x] += treeSize[y];
      if (rank[x] == rank[y]) {
        rank[x]++;
      }
    }

    return true;
  }

  bool same(int x, int y) { return find(x) == find(y); }

  int getSize(int x) { return treeSize[find(x)]; }
};

int main() {
  int n;
  cin >> n;
  vector<pair<pii, int>> xy, yx;
  rep(i, n) {
    int x, y;
    cin >> x >> y;
    xy.eb(mp(mp(x, y), i));
    yx.eb(mp(mp(y, x), i));
  }

  sort(all(xy));
  sort(all(yx));

  vector<pair<int, pii>> cost;
  rep(i, n - 1) {
    cost.eb(mp(xy[i + 1].F.F - xy[i].F.F, mp(xy[i].S, xy[i + 1].S)));
    cost.eb(mp(yx[i + 1].F.F - yx[i].F.F, mp(yx[i].S, yx[i + 1].S)));
  }

  sort(all(cost));

  UnionFindTree uf(n);
  ll ans = 0;
  rep(i, sz(cost)) {
    if (!uf.same(cost[i].S.F, cost[i].S.S)) {
      uf.unite(cost[i].S.F, cost[i].S.S);
      ans += cost[i].F;
    }
  }

  cout << ans << endl;
}