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 LL long long
#define N 100005
using namespace std;
LL ans;
int n,x[N],y[N];
vector<int>g[N];
int fa[N],cnt[N],sz[N];
bool mark[N],mark2[N];
int gf(int k) {
	return fa[k]==k?fa[k]:fa[k]=gf(fa[k]);
}
int main() {
	scanf("%lld",&n);
	for(int i=1; i<N; i++)fa[i]=i;
	for(int i=1; i<=n; i++)scanf("%d%d",&x[i],&y[i]),g[x[i]].push_back(y[i]);
	for(int i=1; i<N; i++) {
		int a;
		if(!g[i].empty())a=g[i][0];
		for(int j=1; j<g[i].size(); j++) {
			int b=g[i][j];
			fa[gf(b)]=gf(a);
		}
	}
	for(int i=1;i<=n;i++){
		sz[gf(y[i])]+=!mark2[y[i]];
		mark2[y[i]]=1;
		cnt[gf(y[i])]+=!mark[x[i]];
		mark[x[i]]=1;
	}
	for(int i=1;i<=n;i++){
		int f=gf(y[i]);
		ans+=1ll*sz[f]*cnt[f];
		sz[f]=0;
	}
	printf("%lld\n",ans-n);
	return 0;
}

