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 ll long long
using namespace std;
int n;
ll ans=0;
const int maxn=1e5+5;
ll a[maxn*2],b[maxn*2],fa[maxn*2];
int find(int x)
{
    return (x==fa[x])?x:(fa[x]=find(fa[x]));
}
int main()
{
    scanf("%d",&n);
    for(int i=0;i<=maxn*2;i++)
        fa[i]=i;
    for(int i=1;i<=n;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        fa[find(x)]=fa[find(y+maxn)];
    }
    for(int i=1;i<=maxn;i++)
    {
        a[find(i)]++;
    }
    for(int i=maxn+1;i<=maxn*2;i++)
    {
        b[find(i)]++;
    }
    for(int i=1;i<=maxn*2;i++)
    {
        ans+=a[i]*b[i];
    }
    if(ans-n<0)
        printf("0\n");
    else
        printf("%lld\n",ans-n);
    return 0;
}
