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 <vector>
#include <iostream>
using namespace std;

vector<vector<int> > adj(200100);
bool done[200100];

void dfs(int i, int &x, int &y, int &z) {
    done[i] = true;
    (i % 2 ? y : x)++;
    for (int j : adj[i]) { z++; if (!done[j]) dfs(j, x, y, z); }
}

int main() {
    int n; cin >> n;
    for (int i = 0; i < n; i++) {
        int a, b; cin >> a >> b;
        adj[2*a].push_back(2*b+1);
        adj[2*b+1].push_back(2*a);
    }
    long long ans = 0;
    for (int i = 0; i <= 200001; i++)
        if (!adj[i].empty() && !done[i]) {
            int x = 0, y = 0, z = 0;
            dfs(i, x, y, z);
            ans += (long long)x * y - z/2;
        }
    cout << ans << endl;
}
