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;
typedef long long ll;
const int MX = 2e5+1, XMX = 1e5+1;
int par[MX], xcnt[MX], ycnt[MX];
int f(int i) {return par[i] == i ? i : par[i] = f(par[i]);}
void join(int i, int j) {
	i = f(i), j = f(j);
	if(i != j) {
		par[j] = i;
		xcnt[i] += xcnt[j];
		ycnt[i] += ycnt[j];
	}
}
int main() {
	ios::sync_with_stdio(0);cin.tie(0);
	int n;
	cin >> n;
	for(int i=1; i<XMX; i++) {
		xcnt[i] = 1;
		par[i] = i;
	}
	for(int i=XMX; i<MX; i++) {
		ycnt[i] = 1;
		par[i] = i;
	}
	for(int i=0; i<n; i++) {
		int x, y;
		cin >> x >> y;
		join(x, y+100000);
	}
	ll tot = 0;
	for(int i=0; i<MX; i++) {
		if(par[i] != i) continue;
		tot += (ll)xcnt[i]*ycnt[i];
	}
	cout << (tot - n);
}