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;
typedef long long ll;
const int maxn=1e5+5;
int n;
int f[maxn<<1];
int sum[maxn<<1][2];
inline int find(int x){
	if(f[x]==x) return x;
	return f[x]=find(f[x]);
}
inline void unionn(int x,int y){
	x=find(x);
	y=find(y);
	if(x!=y) f[y]=x;
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<maxn<<1;i++) f[i]=i;
	for(int i=1;i<=n;i++){
		int x,y;
		scanf("%d%d",&x,&y);
		y+=maxn;
		unionn(x,y);
	}
	for(int i=1;i<maxn<<1;i++) sum[find(i)][i/maxn]++;
	ll ans=0;
	for(int i=1;i<maxn<<1;i++) ans+=(ll)sum[i][0]*sum[i][1];
	printf("%lld\n",ans-n);
	return 0;
}