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 <iomanip>
#include <cstdio>
#include <cmath>
#include <string>
#include <vector>
#include <limits>
#include <algorithm>
#include <utility>
#include <sstream>
#include <numeric>
#include <iterator>

using namespace std;

#define repd(i,a,b) for (int i=(a);i<(b);i++)
#define rep(i,n) repd(i,0,n)
// #define ll long long
typedef long long ll;
typedef pair<int, int> P;

const int V = 100005;
vector<int> to[V*2];
bool visited[V*2];
vector<int> cnt;

void dfs(int v){
  if (visited[v]) return;
  visited[v] = true;
  cnt[v/V]++;
  for (int u : to[v]) dfs(u);
}

int main(int argc, const char * argv[]) {

    int n;
    cin >> n;
    rep(i,n){
      int x, y;
      cin >> x >> y;
      y+= V;
      to[x].push_back(y);
      to[y].push_back(x);
    }
    ll ans = 0;
    rep(i, V*2){
      if (visited[i]) continue;
      cnt = vector<int>(2);
      dfs(i);
      ans += (ll)cnt[0]*cnt[1];
    }
    ans -= n;
    cout << ans << endl;

    return 0;
}
