There are N dots in a two-dimensional plane. The coordinates of the i-th dot are (x_i, y_i).

We will repeat the following operation as long as possible:

* Choose four integers a, b, c, d (a \neq c, b \neq d) such that there are dots at exactly three of the positions (a, b), (a, d), (c, b) and (c, d), and add a dot at the remaining position.



We can prove that we can only do this operation a finite number of times. Find the maximum number of times we can do the operation.

Constraints

* 1 \leq N \leq 10^5
* 1 \leq x_i, y_i \leq 10^5
* If i \neq j, x_i \neq x_j or y_i \neq y_j.
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
x_1 y_1
:
x_N y_N


Output

Print the maximum number of times we can do the operation.

Examples

Input

3
1 1
5 1
5 5


Output

1


Input

2
10 10
20 20


Output

0


Input

9
1 1
2 1
3 1
4 1
5 1
1 2
1 3
1 4
1 5


Output

16
#include <algorithm>
#include <cmath>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <vector>
using namespace std;
// #include "prettyprint.hpp"

const int MAX = 100005;
vector<int> adj[MAX * 2];
bool visited[MAX * 2];
int countX;
int countY;

void dfs(int i) {
  if (visited[i]) return;
  visited[i] = true;

  if (i < MAX) {
    countX++;
  } else {
    countY++;
  }

  // cout << "adj[i]" <<adj[i] << endl;
  for (int j : adj[i]) {
    dfs(j);
  }
}

int main() {
  int N;
  cin >> N;

  for (int i = 0; i < N; i++) {
    int x, y;
    cin >> x >> y;
    adj[x].push_back(y + MAX);
    adj[y + MAX].push_back(x);
  }

  for (int i = 0; i < MAX * 2; i++) visited[i] = false;

  long long ans = 0;
  for (int i = 0; i < MAX * 2; i++) {
    countX = 0;
    countY = 0;
    if (!visited[i]) dfs(i);
    ans += (long long)countX * countY;
  }

  cout << (ans - N) << endl;

  return 0;
}