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;

#define ran 202202

int n;
int x[ran], y[ran];
map<int, int> mpx, mpy;
int p[ran];
int cnt[ran][2];

int get_p(int x) { return p[x]==-1 ? x : p[x] = get_p(p[x]); }

int main() {
	scanf("%d", &n);
	for(int i=1;i<=n;i++) {
		scanf("%d%d", &x[i], &y[i]);
		if (!mpx[x[i]]) mpx[x[i]] = mpx.size();
		x[i] = mpx[x[i]];
		if (!mpy[y[i]]) mpy[y[i]] = mpy.size();
		y[i] = mpy[y[i]];
	}
	for(int i=1;i<=n;i++) y[i] += mpx.size();
	//for(int i=1;i<=n;i++) printf("%d %d\n", x[i], y[i]);
	
	memset(p, -1, sizeof p);
	for(int i=1;i<=n;i++) {
		int a = get_p(x[i]), b = get_p(y[i]);
		if (a!=b) p[a] = b;
	}
	
	for(int i=1;i<=mpx.size() + mpy.size();i++) {
		int a = get_p(i);
		if (i<=mpx.size()) cnt[a][0]++;
		else cnt[a][1]++;
	}
	long long ans = 0;
	for(int i=1;i<=mpx.size() + mpy.size();i++)
		ans += 1LL*cnt[i][0] * cnt[i][1];
	cout<<ans-n<<endl;
	
	return 0;
}