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>
#define debug(x) cerr << #x << ": " << x << '\n';
using namespace std;
using ll = long long;
using P = pair<int, int>;
const int INF = (int)1e9;
const int MOD = (int)1e9 + 7;

const int V = 100001;
vector<int> g[V*2];
int used[V*2];
int cnt[2];

void dfs(int v){
	if(used[v]) return;
	used[v] = true;

	cnt[v/V]++;
	for(int nv: g[v]){
		if(used[nv]) continue;
		dfs(nv);
	}
}

int main(void){
	int N;
	cin >> N;
	for(int i = 0; i < N; i++){
		int x, y;
		cin >> x >> y;
		y += V;
		g[x].push_back(y);
		g[y].push_back(x);
	}

	ll res = 0;
	for(int i = 0; i < V*2; i++){
		if(used[i]) continue;

		cnt[0] = 0;
		cnt[1] = 0;
		dfs(i);
		res += (ll)cnt[0]*cnt[1];
	}

	res -= N;

	cout << res << '\n';

	return 0;
}