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 IO_OP std::ios::sync_with_stdio(0); std::cin.tie(0);
#define F first
#define S second
#define PB push_back
#define MP make_pair
#define EB emplace_back

using namespace std;

typedef long long ll;
typedef pair<int, int> pi;
typedef vector<int> vi;

const int INF = 1e9 + 7, V = 1e5 + 87;

vi G[V*2];
int l, r, vis[2*V];

void dfs(int u) {
	vis[u] = 1;
	for(int v:G[u])
		if(!vis[v])
			dfs(v);
	if(u < V) l++;
	else r++;
}

int main()
{
	IO_OP;
	
	int n;
	cin >> n;
	for(int i=0;i<n;i++) {
		int x, y;
		cin >> x >> y;
		y += V;
		G[x].PB(y);
		G[y].PB(x);
	}
	ll ans = 0;
	for(int u=1;u<=V*2;u++) {
		if(!vis[u]) {
			l = r = 0;
			dfs(u);
			ans += (ll) l * r;
		}
	}
	cout << ans - n << endl;
}

