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;
const int N=2e6+5;
inline int read(){
    int x=0; char c=getchar();
    while(c<'0'||c>'9')c=getchar();
    while(c>='0'&&c<='9')x=(x<<3)+(x<<1)+(c^48),c=getchar();
    return x;
}
int x[N],y[N],f[N];
set<int>sx[N],sy[N];
int find(int x){return f[x]==x?x:f[x]=find(f[x]);}
int main(){
	int n=read();
	for(int i=0;i<n;i++)x[i]=read(),y[i]=read();
	for(int i=0;i<N;i++)f[i]=i;
	for(int i=0;i<n;i++)f[find(x[i])]=find(y[i]+N/2);
	for(int i=0;i<n;i++){
		int fa=find(x[i]);
		sx[fa].insert(x[i]);
		sy[fa].insert(y[i]);
	}
	long long ans = 0;
	for(int i=0;i<N;i++)ans+=1ll*sx[i].size()*sy[i].size();
	printf("%lld",ans-n);
	return 0;
}