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;
#define int long long
#define fi first
#define se second
int n,a[200005], b[200005];
int cur = 1, tr[200005], dc[200005];
int c = 100000;
pair<int,int> store[200005];
int f(int u){
	return (a[u] < 0 ? u : a[u] = f(a[u]));
}
void m(int u,int v){
	u = f(u); v = f(v);
	if(u == v) return;
	if(a[u] > a[v]) swap(u,v);
	a[u] += a[v]; a[v] = u;
}
signed main(){
	cin >> n;
	for(int i = 1 ; i <= 200004 ; i ++) a[i] = -1;
	for(int i = 1 ; i <= n ; i ++){
		int x, y;
		cin >> x >> y;
		b[x] = 1; b[c + y] = 1;
		m(x, c + y);
	}
	int ans = 0;
	for(int i = 1 ; i <= 200004 ; i ++){
		if(b[i]){
			if(!dc[f(i)]){
				dc[f(i)] = 1;
				tr[f(i)] = cur;
				cur ++;
			}
			if(i <= c) store[tr[f(i)]].fi ++;
			else store[tr[f(i)]].se ++;
		}
	}
	for(int i = 1 ; i < cur ; i ++)
		ans += store[i].fi * store[i].se;
	cout << ans - n;
}
