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 <bits/stdc++.h>
using namespace std;

#define int long long

const int N = 1e5 + 10;
int n, mark[2][N], cnt[2][N];
vector <int> adj[2][N];

void dfs(int x, int y, int c) {
	mark[x][y] = c;
	cnt[x][c]++;
	for (int u : adj[x][y])
		if (!mark[x ^ 1][u])
			dfs(x ^ 1, u, c);
}

int32_t main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n;
	for (int i = 0; i < n; i++) {
		int x, y;
		cin >> x >> y;
		x--; y--;
		adj[0][x].push_back(y);
		adj[1][y].push_back(x);
	}
	
	for (int i = 0, x = 1; i < N; i++)
		if (!mark[0][i])
			dfs(0, i, x++);
	
	int ans = -n;
	for (int i = 0; i < N; i++)
		ans += cnt[0][i] * cnt[1][i];
	cout << ans << endl;
}