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 <vector>
#define repi(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,a) repi(i,0,a)
#define all(a) (a).begin(), (a).end()

using ll = long long;

int N;
std::vector<int> G[200010];
bool used[200010];
std::vector<int> us, vs;
ll ans = 0;

void dfs( int v )
{
  used[v] = true;

  (v>100000?vs:us).push_back( v );

  for( auto &u : G[v] )
  {
    if( !used[u] )
      dfs( u );
  }
}

int main()
{
  scanf( "%d", &N );
  rep( i, N )
  {
    int x, y;
    scanf( "%d%d", &x, &y );
    --x; --y;
    y += 100001;

    G[x].push_back( y );
    G[y].push_back( x );
  }

  rep( i, 100000 )
  {
    us.clear();
    vs.clear();

    if( !used[i] )
    {
      dfs( i );

      ans += us.size()*vs.size();
    }
  }

  printf( "%lld\n", ans-N );

  return 0;
}