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;

//const long double PIL = 3.141592653589793238462643383279502884L;
//const double PI = 3.14159265358979323846;

#define sz(a) int((a).size())
#define all(c) (c).begin(), (c).end()

using ll = long long;
using ii = pair<int,int>;
using vi = vector<int>;
using vll = vector<long long>;
using vii = vector<pair<int,int>>;

const int N = 1e5;
vi adj[2*N+1];
bool visited[2*N+1];
int u, v;

void dfs(int x) {
	visited[x] = true;
	if (x <= N) ++u;
	else ++v;
	for (int y : adj[x])
		if (!visited[y]) dfs(y);
}

int main() {
	//~ ios::sync_with_stdio(0);
	//~ cin.tie(0);
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		int x, y;
		scanf("%d%d", &x, &y);
		y += N;
		adj[x].push_back(y);
		adj[y].push_back(x);
	}

	ll answer = 0;
	for (int x = 1; x <= N; x++) {
		if (visited[x]) continue;
		u = 0, v = 0;
		dfs(x);
		answer += (ll)u * v;
	}
	printf("%lld\n", answer - n);
}
