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;

vector<int> AdjList[200015];

bool visited[200015];

int main(){
	int N;
	scanf("%d", &N);
	for(int i = 0; i < N; i ++){
		int x, y;
		scanf("%d%d", &x, &y);
		AdjList[x].push_back(100005+y);
		AdjList[100005+y].push_back(x);
	}
	
	memset(visited, 0, sizeof(visited));
	long long ans = 0;
	for(int i = 0; i < 200015; i ++){
		if(!visited[i]){
			visited[i] = 1;
			long long X = 0;
			long long Y = 0;
			
			long long edges = AdjList[i].size();
			if(i < 100003){
				X ++;
			}else{
				Y ++;
			}
			queue<int> q;
			q.push(i);
			while(!q.empty()){
				int u = q.front(); q.pop();
				for(int v: AdjList[u]){
					if(!visited[v]){
						visited[v] = 1;
						q.push(v);
						edges += AdjList[v].size();
						if(v < 100003){
							X ++;
						}else{
							Y ++;
						}
					}
				}
			}
			ans += X*Y-(edges/2);
			//printf("%lld %lld %lld\n", X, Y, edges/2);
		}
	}
	printf("%lld", ans);
}
