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 <iostream>
#include <bits/stdc++.h>

using namespace std;
bool vis[200005];
vector<int> rad[200005];
int n;
int xcnt,ycnt;
void dfs(int now)
{
    vis[now] = true;
    if(now>100000) ++xcnt;
    else ++ycnt;
    for(int i=0;i<rad[now].size();++i)
    {
        if(vis[rad[now][i]]) continue;
        dfs(rad[now][i]);
    }
}

int main()
{
    scanf("%d",&n);
    int u,v;
    for(int i=0;i<n;++i)
    {
        scanf("%d%d",&u,&v);
        u+=100000;
        rad[u].push_back(v);
        rad[v].push_back(u);
    }
    long long ecnt = 0;
    for(int i=1;i<=100000;++i)
    {
        if(vis[i]) continue;
        xcnt = 0;ycnt = 0;
        dfs(i);
        ecnt += 1LL * xcnt * ycnt;
    }
    printf("%lld\n",ecnt-n);
    return 0;
}
