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>
#define Rint register int
using namespace std;
typedef long long LL;
const int N = 200003;
template<typename T>
inline void read(T &x){
	int ch = getchar(); x = 0;
	bool flag = false;
	while((ch < '0' || ch > '9') && ch != '-') ch = getchar();
	if(ch == '-'){
		flag = true;
		ch = getchar();
	}
	while(ch >= '0' && ch <= '9'){
		x = x * 10 + ch - '0';
		ch = getchar();
	}
	if(flag) x = -x;
}
int n, fa[N], siz[N][2];
LL ans;
inline int getfa(int x){
	return x == fa[x] ? x : fa[x] = getfa(fa[x]);
}
int main(){
	read(n);
	for(Rint i = 1;i <= 200000;i ++) fa[i] = i, siz[i][0] = (i <= 100000), siz[i][1] = (i > 100000);
	for(Rint i = 1;i <= n;i ++){
		int x, y;
		read(x); read(y); y += 100000;
		x = getfa(x); y = getfa(y);
		if(x != y){
			ans += (LL) siz[x][0] * siz[y][1] + (LL) siz[x][1] * siz[y][0];
			siz[x][0] += siz[y][0];
			siz[x][1] += siz[y][1];
			fa[y] = x;
		}
	}
	printf("%lld\n", ans - n);
}