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
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Main main = new Main();
        main.solve();
    }
    public void solve() {
        Scanner scan = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        int N = scan.nextInt();
        int[] x = new int[N];
        int[] y = new int[N];
        HashMap<Integer, List<Integer>> edges = new HashMap<>();
        for (int i = 0; i < N; i++) {
            x[i] = scan.nextInt();
            y[i] = scan.nextInt()+100005;
            if (!edges.containsKey(x[i])) {
                edges.put(x[i], new ArrayList<>());
            }
            if (!edges.containsKey(y[i])) {
                edges.put(y[i], new ArrayList<>());
            }
            edges.get(x[i]).add(y[i]);
            edges.get(y[i]).add(x[i]);
        }
        boolean[] visited = new boolean[100005 * 2];
        long count = 0;
        for (int i = 0; i < N; i++) {
            if (visited[x[i]]) {
                continue;
            }
            long[] cnt = new long[2];
            cnt[0] = 0;//x
            cnt[1] = 0;//y
            dfs(cnt, visited, edges, x[i]);
            count += cnt[0] * cnt[1];
        }
        System.out.println(count - N);
    }
    public void dfs(long[] cnt, boolean[] visited, HashMap<Integer, List<Integer>> edges, int x) {
        if (visited[x]) {
            return;
        }
        visited[x] = true;
        if (x < 100005) {
            cnt[0] += 1;
        } else {
            cnt[1] += 1;
        }
        for (int y : edges.get(x)) {
            dfs(cnt, visited, edges, y);
        }
    }
}
