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;
typedef long double ld;

bool visited[200005];
vector <int> graf[2000005];

ll sa, sb, br;

void dfs(int v){
    visited[v] = 1;
    if(v <= 100000) sa++;
    else sb++;
    for(auto c : graf[v]){
        br++;
        if(visited[c]) continue;
        dfs(c);
    }
}

ll res;

int main(){
    ios_base::sync_with_stdio(false);
	cout.precision(10);
	cout << fixed;

    int n;
    cin >> n;
    for(int i=1; i<=n; i++){
        int a, b;
        cin >> a >> b;
        graf[a].push_back(b+100000);
        graf[b+100000].push_back(a);
    }
    for(int i=1; i<=200000; i++){
        if(!visited[i]){
            sa = 0;
            sb = 0;
            br = 0;
            dfs(i);
            br /= 2;
            res += sa*sb-br;
        }
    }
    cout << res;
    return 0;
}
