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
/*
Task:F - Must Be Rectangular!
Statement: Given N points (1<=x,y<=10^5)
You can do the following operation find acute triangle
add the point to form it rectangular if that point doesn't exist yet;
Find the maximum number of operation
*/
#include<bits/stdc++.h>
using namespace std;
const int maxn=2e5+100;
int n,u,v,vis[maxn],x,y;
vector<int> a[maxn];
void dfs(int nod,int dis){
    vis[nod]=true;
    if(dis&1) x++;
    else y++;
    for(auto i:a[nod]){
        if(!vis[i]){
            dfs(i,dis+1);
        }
    }
    return;
}

int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d%d",&u,&v);
        v+=(100000);
        a[u].push_back(v);
        a[v].push_back(u);
    }
    long long ans=0;
    for(int i=1;i<=200010;i++){
        if(!vis[i]){
            x=0,y=0;
            dfs(i,0);
            ans+=(1LL*x*y);
        }
    }
    printf("%lld\n",ans-n);
}
/*
6
1 1
1 2
1 3
2 2
2 3
2 4
*/
