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
#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
#define int long long
#define F first
#define S second
#define P pair
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define rep(i,a,b) for(int i=a;i<b;i++)
#define V vector
#define RE return
#define ALL(a) a.begin(),a.end()
#define MP make_pair
#define PB push_back
#define PF push_front
#define FILL(a,b) memset(a,b,sizeof(a))
using namespace std;
int fa[200015],n,p[200015][2],maxi=1e5;
int get(int x){
    if(fa[x]==x)RE x;
    RE fa[x]=get(fa[x]);
}
void up(int x,int y){
    x=get(x);
    y=get(y);
    if(x!=y){
        fa[y]=x;
    }
}
signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin>>n;
    int x,y;
    FOR(i,1,2*maxi)fa[i]=i;
    FOR(i,1,n){
        cin>>x>>y;
        up(x,y+maxi);
    }
    int ans=0;
    FOR(i,1,maxi)p[get(i)][0]++;
    FOR(i,maxi+1,maxi*2)p[get(i)][1]++;
    FOR(i,1,2*maxi){
        ans=ans+p[i][0]*p[i][1];
    }
    cout<<ans-n;
	RE 0;
}

