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;

typedef long long ll;
const int N = 1e5 + 100;

int n, x, y;
ll ans;
bool mark[N];
vector<int> nei[N], have[N], vec;
set<int> st[N];

void add(int u, int v) {
	nei[u].push_back(v);
	nei[v].push_back(u);
}

void dfs(int v) {
	mark[v] = true;
	vec.push_back(v);
	for (int u: nei[v])
		if(mark[u] == false)
			dfs(u);
}

int main() {
	ios::sync_with_stdio(false), cin.tie(0);
	cin >> n;
	ans = -n;
	while(n--) {
		cin >> x >> y;
		have[y].push_back(x);
		st[x].insert(y);
	}
	for (int i = 0; i < N; i++)
		for (int j = 1; j < have[i].size(); j++)
			add(have[i][j - 1], have[i][j]);
	for (int i = 0; i < N; i++)
		if(mark[i] == false) {
			vec.clear();
			dfs(i);
			set<int> help;
			for (int u: vec)
				help.insert(st[u].begin(), st[u].end());
			ans += 1LL * vec.size() * help.size();
		}
	cout << ans;
	return 0;
}