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;
typedef long long LL;
int a[100001], b[100001];
int p[200001];
vector<int> v[200001];
set<int> r, c;
int root(int a){
    return p[a] == a?a:(p[a]=root(p[a]));
}
void merge(int a, int b){
    a = root(a), b = root(b);
    if(a != b) p[a] = b;
}
int main(){
    int n;
    scanf("%d", &n);
    for(int i=1;i<=200000;i++) p[i] = i;
    for(int i=1;i<=n;i++){
        scanf("%d%d", &a[i], &b[i]);
        merge(a[i], 100000+b[i]);
    }
    for(int i=1;i<=n;i++){
        v[root(a[i])].push_back(i);
    }
    LL ans = 0;
    for(int i=1;i<=200000;i++){
        if(root(i) == i){
            r.clear(), c.clear();
            for(int j=0;j<v[i].size();j++){
                int k = v[i][j];
                int y = a[k], x = b[k];
                r.insert(y);
                c.insert(x);
            }
            ans += r.size()*1ll*c.size() - v[i].size();
        }
    }
    printf("%lld\n", ans);
}