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 fi first
#define se second
#define mp make_pair
#define pb push_back

using namespace std;

LL gcd(LL a,LL b){return b?gcd(b,a%b):a;}
LL lcm(LL a,LL b){return a/gcd(a,b)*b;}
LL powmod(LL a,LL b,LL MOD){LL ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
const int N = 100005;
int a[N<<2],b[N<<2],f[N<<2],n;
int find(int x){return x==f[x]?x:f[x]=find(f[x]);}
int main(){
	ios::sync_with_stdio(false);
	cin>>n;
	for(int i=0;i<=2*N;i++)f[i]=i;
	for(int i=1;i<=n;i++){
		int x,y;
		cin>>x>>y;
		f[find(x)]=find(y+N);
	}		
	for(int i=1;i<=N;i++)a[find(i)]++;
	for(int i=N+1;i<=2*N;i++)b[find(i)]++;
	LL ans=0;
	for(int i=1;i<=2*N;i++)ans+=1ll*a[i]*b[i];
	cout<<ans-n<<endl;
	return 0;
}