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 <cstdio>
#include <cstring>
#define maxn 100010
#define ll long long

struct node{int x,y,next;};
node e[4*maxn];
int first[2*maxn];
void buildroad(int x,int y)
{
	static int len=0;
	e[++len]=(node){x,y,first[x]};
	first[x]=len;
}
int n,m=0;
bool v[maxn*2],vis[maxn*2];
void dfs_getm(int x)
{
	if(x>maxn-10)m++;v[x]=true;
	for(int i=first[x];i;i=e[i].next)
	if(!v[e[i].y])dfs_getm(e[i].y);
}
ll ans=0;
void dfs_collectans(int x)
{
	if(x<=maxn-10)
	{
		int tot=0;
		for(int i=first[x];i;i=e[i].next)
		tot++;
		ans+=(ll)m-tot;
	}
	vis[x]=true;
	for(int i=first[x];i;i=e[i].next)
	if(!vis[e[i].y])dfs_collectans(e[i].y);
}

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		int x,y;
		scanf("%d %d",&x,&y);
		buildroad(x,y+maxn-10);
		buildroad(y+maxn-10,x);
	}
	for(int i=1;i<=maxn-10;i++)
	{
		if(first[i]==0||v[i])continue;
		m=0;
		dfs_getm(i);
		dfs_collectans(i);
	}
	printf("%lld",ans);
}