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<iostream>
#include<map>
#include<algorithm>
#include<functional>
#include<bitset>
#include<vector>
#include<set>
using  namespace std;
#define ll long long
#define pb push_back
#define sc second
#define fr first
#define mp make_pair
const int  MOD = 1e9 + 7, OO = 1e9;
const int  N = 1e5 + 10;
int vis[N][3];
vector<int> g[3][N];
set<int> st[2];
void dfs(int u, int t) {
	if (vis[u][t])return;
	vis[u][t] = 1;
	st[t].insert(u);
	for (int i = 0; i < g[t][u].size(); ++i) {
		dfs(g[t][u][i], !t);
	}
}
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int n;
	cin >> n;
	for (int i = 1; i <= n; ++i) {
		int a, b;
		cin >> a >> b;
		g[0][a].pb(b);
		g[1][b].pb(a);
	}
	ll ans = 0;
	for (int it = 0; it < 2; ++it) {
		for (int i = 1; i < N; ++i) {
			dfs(i, it);
			ans += st[0].size()*1ll*st[1].size();
			st[1].clear();
			st[0].clear();
		}
	}
	cout << ans - n << endl;

}