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 max_n=200005;
int n,par[max_n];
int findpar(int x){
    return x==par[x]?x:par[x]=findpar(par[x]);
}
int sx[max_n],sy[max_n];
void unite(int x,int y){
    x=findpar(x);y=findpar(y);
    if(x!=y){
        par[x]=y;
        sx[y]+=sx[x],sy[y]+=sy[x];
    }
}
int main(){
    scanf("%d",&n);
    for(int i=1;i<max_n;i++){
        par[i]=i;
        if(i<=100000)sx[i]=1;
        else sy[i]=1;
    }
    for(int i=1;i<=n;i++){
        int x,y;
        scanf("%d%d",&x,&y);
        unite(x,y+100000);
    }
    ll ans=0;
    for(int i=1;i<max_n;i++){
        if(findpar(i)==i){
            ans+=1ll*sx[i]*sy[i];
        }
    }
    ans-=n;
    printf("%lld\n",ans);
    return 0;
}