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 lim = 1e5+5;
int n,deg[lim],vis[2][lim];
vector<int> adj[2][lim];
void dfs(int x,int s,vector<int>& l,vector<int>& r) {
	if(vis[s][x]==1)
		return;
	vis[s][x] = 1;
	if(s==0)
		l.push_back(x);
	else
		r.push_back(x);
	for(auto y:adj[s][x])
		dfs(y,s^1,l,r);
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    cin>>n;
    for(int i=0;i<n;i++) {
    	int x,y;
    	cin>>x>>y;
    	adj[0][x].push_back(y);
    	adj[1][y].push_back(x);
    	deg[x]++;
    }

    ll ans = 0;
    for(int i=1;i<=100000;i++) {
    	vector<int> l,r;
    	dfs(i,0,l,r);
    	int sum = 0;
    	for(auto x:l)
    		sum += deg[x];
    	ans += l.size()*1LL*r.size() - sum;
    }
    cout<<ans;

	return 0;
}