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
/*In the name of Allah*/
#include <bits/stdc++.h>
using namespace std;
#define inf 1000000111
#define MAX 200005
#define mod 1000000007
#define ll long long
vector<int> g[MAX];
set<int> temp;
bool see[MAX];
int x[MAX],y[MAX];
vector<int> m[MAX],mm[MAX],yy;
set<int> xx;
int cnt = 0;
void dfs(int u){
	see[u] = 1;
	cnt++;
	for(auto i : m[u])temp.insert(i);
	for(auto v : g[u])
		if(!see[v])
			dfs(v);
}


int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	int n;
	cin>>n;
	for(int i = 0; i < n ; i++)cin>>x[i]>>y[i],m[x[i]].push_back(y[i]),mm[y[i]].push_back(x[i]),yy.push_back(y[i]),xx.insert(x[i]);
	for(auto v : yy){
		for(int i = 1; i < mm[v].size() ; i++){
			g[mm[v][i]].push_back(mm[v][i-1]);
			g[mm[v][i-1]].push_back(mm[v][i]);
		}
	}	
	ll ans = 0;
	for(auto i : xx)
		if(!see[i])cnt = 0,temp.clear(),dfs(i),ans += 1LL * temp.size() * cnt;
	cout<<ans - n;
	return 0;
}
