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;
const int maxn = 234566;
int u[maxn];
int find(int x){
	if(x == u[x]) return x;
	return u[x] = find(u[x]);
}
int x[maxn], y[maxn];
set<int> xx[maxn], yy[maxn];
int main(){
	int n; cin >> n;
	for(int i = 0; i < n; i++) cin >> x[i] >> y[i];
	iota(u, u + maxn, 0);
	for(int i = 0; i < n; i++) u[find(x[i])] = find(y[i] + maxn / 2);
	for(int i = 0; i < n; i++){
		int nd = find(x[i]);
		xx[nd].insert(x[i]);
		yy[nd].insert(y[i]);
	}
	long long ans = 0;
	for(int i = 0; i < maxn; i++){
		ans += 1ll * xx[i].size() * yy[i].size();
	}
	cout << ans - n << endl;
	return 0;
}
