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 mp make_pair
#define pb push_back
#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)
#define debug(x) cerr << #x << " " << x << '\n'
using namespace std;
using ll = long long;
using pii = pair<int,int>;
using pli = pair<ll,int>;
const int INF = 0x3f3f3f3f, N = 1e5 + 5;
const ll LINF = 1e18 + 5;
int n, x[N], y[N], f[N<<1], sz[N<<2][2];
int find(int x)
{
	if(x==f[x]) return x;
	return f[x] = find(f[x]);
}
int main()
{
 	ios::sync_with_stdio(false);
 	cin.tie(0);
 	cin >> n;
 	for(int i=1; i<N*2; i++) f[i] = i;
 	for(int i=1; i<=n; i++) 
 	{
 		cin >> x[i] >> y[i];
 		f[find(x[i])] = find(y[i]+N);
	}
	for(int i=1; i<N*2; i++) sz[find(i)][i/N]++;
	ll ans = 0;
	for(int i=1; i<N*2; i++) 
	{
		ans += 1ll*sz[i][0]*sz[i][1];
	}
	ans -= n;
	cout << ans;
	return 0;
}
