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 <vector>
using namespace std;
int const s = 100000;
vector< vector<int> > edg(s*2);
vector<bool> vsd(s*2, false);
void f(int v, int& x, int& y){
    vsd[v] = true;
    if(v < s)x++;
    else y++;
    for(int i=0;i<edg[v].size();i++){
        int w = edg[v][i];
        if(!vsd[w]) f(w, x, y);
    }
}
int main() {
    int n;
    long long ans = 0;
    cin >> n;
    for(int i=0;i<n;i++){
        int x, y;
        cin >> x >> y;
        x--;
        y += s-1;
        edg[x].push_back(y);
        edg[y].push_back(x);
    }
    for(int i=0;i<s*2;i++){
        if(vsd[i])continue;
        int x = 0, y = 0;
        f(i, x, y);
        ans += (long long) x*y;
    }
    ans -= n;
    cout << ans << endl;
	return 0;
}