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 <cstdio>
#include <vector>
#define N 100000
using namespace std;
typedef long long ll;

ll n, x, y, ans, cx, cy, v[200005];
vector<ll> gr[200005];

void dfs(int p) {
	v[p] = 1;
	p<=N ? cx++ : cy++;
	for(int i=0; i<gr[p].size(); i++) {
		if(!v[gr[p][i]]) dfs(gr[p][i]);
	}
}

int main() {
	ll i, x, y;
	cin >> n;
	for(i=0; i<n; i++) {
		scanf("%lld%lld", &x, &y);
		gr[x].push_back(y+N);
		gr[y+N].push_back(x);
	}
	for(i=0; i<=2*N; i++) {
		if(!v[i]) {
			cx = cy = 0;
			dfs(i);
			ans += cx * cy;
		}
	}
	cout << ans - n;
	return 0;
}