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>
typedef long long int ll;
typedef unsigned long long int ull;
#define BIG_NUM 2000000000
#define HUGE_NUM 1000000000000000000
#define MOD 1000000007
#define EPS 0.000000001
using namespace std;




#define SIZE 100005

ll N;
ll num_X,num_Y;
bool visited[2*SIZE];
vector<int> G[2*SIZE];


void dfs(int node_id){

	for(int i = 0; i < G[node_id].size(); i++){

		int next = G[node_id][i];
		if(visited[next])continue;

		visited[next] = true;
		if(next >= SIZE){

			num_Y++;
		}else{

			num_X++;
		}
		dfs(next);
	}
}


int main(){

	scanf("%lld",&N);

	ll x,y;

	for(ll i = 0; i < N; i++){

		scanf("%lld %lld",&x,&y);
		y += SIZE;

		G[x].push_back(y);
		G[y].push_back(x);
	}

	for(int i = 0; i < 2*SIZE; i++){

		visited[i] = false;
	}

	ll ans = 0;

	for(int i = 0; i < SIZE; i++){
		if(visited[i])continue;

		visited[i] = true;
		num_X = 1;
		num_Y = 0;

		dfs(i);

		ans += num_X*num_Y;
	}

	ans -= N;

	printf("%lld\n",ans);

	return 0;
}
