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 LL long long
#define N 200005
#define M 100000
bool mark[N];
int xi[N],E,L,R;
vector <int> G[N];
void dfs(int now,bool l) {
	E ++;
	if(mark[now]) return ;
	mark[now] = 1;
	if(l) L ++;
	else R ++;
	for(int i=0; i<G[now].size(); i++) {
		int e = G[now][i];
		dfs(e,!l);
	}
	return ;
}
int main () {
	int n;
	scanf("%d",&n);
	for(int i=1,x,y; i<=n; i++) {
		scanf("%d%d",&x,&y);
		G[x].push_back(y+M),G[y+M].push_back(x);
	}
	LL ans = 0;
	for(int i=1; i<=M; i++) {
		if(!mark[i]) {
			E = -1,L = 0,R = 0;
			dfs(i,1);
			if(L < 2 || R < 2) continue;
			ans += 1LL*L*R - E/2;
		}
	}
	cout<<ans;
	return 0;
}