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;
#define ll long long
#define pb push_back
ll x,y;
bool vis[200001];
vector<int> adj[200001];
void dfs(int s)
{
    if(s<=1e5) x++;
    else y++;
    for(auto v: adj[s])
    {
    if(!vis[v])
    {
        vis[v]=true;
        dfs(v);
    }
    }
}
int main()
{
    ll i,n,res=0;
    cin>>n;
    for(i=0;i<n;i++)
    {
        int u,v;
        cin>>u>>v;
        adj[u].pb(100000+v);
        adj[100000+v].pb(u);
    }
   
    
   
    for(i=0;i<200001;i++)
    {
        if(!vis[i] && !adj[i].empty())
        {
            x=0;y=0;
            vis[i]=true;
            dfs(i);
            res+=(x*y);
        }
    }
    cout<<res-n;
    return 0;
}