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 rep(i, n) for(int (i)=0; (i)<(n); (i)++)
using namespace std;

int N;
const int MAX_N = 100005;
const int MAX_V = 100005;
vector<int> edges[MAX_N * 2];
bool visited[MAX_N * 2];

void dfs(int v, vector<int> &cnt) {
  if(visited[v]) return;
  visited[v] = true;
  cnt[v/MAX_V]++;
  for(int u : edges[v]) dfs(u, cnt);
}

int main() {
  cin >> N;
  rep(i, N) {
    int x, y;
    cin >> x >> y;
    edges[x].push_back(y+MAX_V);
    edges[y+MAX_V].push_back(x);
  }
  long long ans=0;
  rep(i, MAX_N*2) {
    if(visited[i]) continue;
    vector<int> cnt(2);
    dfs(i, cnt);
    ans += (long long)cnt[0]*cnt[1];
  }
  cout << ans-N << endl;
  return 0;
}

