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 <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <cmath>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>
#include <complex>
#include <unordered_map>
#include <unordered_set>
#include <random>
#include <cassert>
#include <fstream>
#define popcount __builtin_popcount
using namespace std;
typedef long long int ll;
typedef pair<int, int> P;
vector<int> g[200020];
bool used[200020];
ll cnt1, cnt2;
void dfs(int x){
	used[x]=1;
	if(x<100000) cnt1++;
	else cnt2++;
	for(auto y:g[x]){
		if(!used[y]) dfs(y);
	}
}
int main()
{
    int n; cin>>n;
	for(int i=0; i<n; i++){
		int x, y; cin>>x>>y; x--; y--;
		g[x].push_back(y+100000);
		g[y+100000].push_back(x);
	}
	ll ans=0;
	for(int i=0; i<200000; i++){
		if(!used[i]){
			cnt1=0, cnt2=0;
			dfs(i);
			ans+=cnt1*cnt2;
		}
	}
	ans-=(ll)n;
	cout<<ans<<endl;
    return 0;
}