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;
const int MAX_N=100010;
typedef long long ll;

vector<int> G[200200];
bool used[200200];

int main(){
    int N; cin >> N;
    for(int i=0; i<N; i++){
        int x, y; cin >> x >> y;
        x--; y--; y+=MAX_N;
        G[x].push_back(y);
        G[y].push_back(x);
    }
    //じゅんびおけ

    ll num=0;

    for(int i=0; i<2*MAX_N; i++){
        if(used[i]) continue;
        queue<int> que;
        que.push(i);
//        used[i]=true;
        ll numx=0, numy=0;
        while(!que.empty()){
            int v=que.front(); que.pop(); used[v]=true;
            if(v<MAX_N) numx++; else numy++;
            for(auto x : G[v]){
                if(!used[x]) {
                    que.push(x);
                    used[x]=true;
                }
            }
        }
        num +=numx*numy;
    }

    cout << num-(ll)N << endl;
}