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 int MAXn = 1e5 + 10;

vector<int> x[MAXn], y[MAXn];
long long sumdeg, n, cntup, cntdown, ans;
bool flag[MAXn][2];
void dfs(int a, bool down) {
	if (down) {
		cntdown++;
		flag[a][1] = true;
		for (int i : y[a]) {
			if (!flag[i][0])
				dfs(i, 0);
		}
	}
	else {
		cntup++;
		sumdeg += x[a].size();
		flag[a][0] = true;
		for (int i : x[a]) {
			if (!flag[i][1])
				dfs(i, 1);
		}
	}
}


int main() {
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	cin >> n;
	for (int i = 0; i < n; i++) {
		int a, b;
		cin >> a >> b;
		a--;
		b--;
		x[a].push_back(b);
		y[b].push_back(a);
	}
	for (int i = 0; i < MAXn; i++)
		if (!flag[i][0]) {
			dfs(i, 0);
			ans += (cntdown * cntup) - sumdeg;
			cntdown = cntup = sumdeg = 0;
		}
	cout << ans;
}
