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>

using namespace std;

int N;
int par[200005];
long long sz1[200005], sz2[200005];

int getrt(int n){
    return par[n] = (par[n] == n ? n : getrt(par[n]));
}

int main(){
    cin.sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> N;
    for(int i = 1; i<=200000; i++){
        par[i] = i;
        if(i <= 100000){
            sz1[i]++;
        }
        else{
            sz2[i]++;
        }
    }
    for(int i =1 ; i<=N; i++){
        int x, y;
        cin >> x >> y;
        y += 100000;
        int px = getrt(x), py = getrt(y);
        if(px != py){
            par[px] = py;
            sz1[py] += sz1[px];
            sz2[py] += sz2[px];
        }
    }
    long long ans = -N;
    for(int i = 1; i<=200000; i++){
        if(par[i] == i){
            ans += sz1[i] * sz2[i];
        }
    }
    cout << ans << endl;
}
