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;
 
vector<vector<int>> graph(2*1e5);
vector<bool> visited(2*1e5);
vector<long long int> num(2);
 
void DFS(int v){
    if(visited[v] == true) return ;
 
    visited[v] = true;
    if(v < 1e5) num[0] += 1;
    else        num[1] += 1;
    for(int next_v : graph[v]) DFS(next_v);
}
 
int main(){
    int n;
    cin >> n;
 
    for(int i=0; i<n; i++){
        int x, y;
        cin >> x >> y;
        x -= 1;
        y -= 1;
        y += 1e5;
        graph[x].push_back(y);
        graph[y].push_back(x);
    }
 
    long long int ans = 0;
    for(int v=0; v<1e5; v++){
        if(visited[v] == true) continue;
        num.assign(2, 0);
        DFS(v);
        ans += num[0] * num[1];
    }
    cout << ans - n << endl;
    return 0;
}
