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
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 200005
#define inf 1e9
#define eps 1e-10
using namespace std;

inline int read()
{
	int x=0,w=1; char c=getchar();
	while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
	while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
	return w==1?x:-x;
}

ll n,vis[maxn],c1,c2,ans;
vector <int> mp[maxn];

inline void dfs(int u)
{
	vis[u]=1; if(u<=100000) c1++; else c2++;
	for(int i=0;i<mp[u].size();i++)
	{
		int v=mp[u][i];
		if(!vis[v]) dfs(v);
	}
}

int main()
{
	n=read();
	for(int i=1;i<=n;i++)
	{
		int x=read(),y=read();
		mp[x].push_back(y+100000);
		mp[y+100000].push_back(x);
	}
	for(int i=1;i<=100000;i++) if(!vis[i]) c1=0,c2=0,dfs(i),ans+=c1*c2;
	cout<<ans-n<<endl;
	return 0;
}