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 <iostream>
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll fa[200001],ra[200001],n,a,b,ans;
map<ll,ll>mx,my;
ll find(ll x){
    if(x==fa[x])return fa[x];
    return fa[x]=find(fa[x]);
}
void uni(ll x,ll y){
    x=find(x),y=find(y);
    if(x==y)return ;
    if(ra[x]<ra[y])swap(x,y);
    fa[y]=x;
    if(ra[x]==ra[y])ra[x]++;
}
int main(void){
    cin>>n;
    for(int i=0;i<200001;i++)fa[i]=i;
    for(int i=0;i<n;i++){
        cin>>a>>b;
        uni(a,b+100000);
    }
    for(int i=1;i<=100000;i++)mx[find(i)]++;
    for(int i=100001;i<=200000;i++)my[find(i)]++;
    for(int i=1;i<=100000;i++)ans+=mx[i]*my[i];
    cout<<ans-n<<endl;
}
