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 V = 100001;

int N, x[V], y[V];
vector<int> G[V * 2];
bool vis[V*2];
vector<int> cnt(2);

void dfs(int v){
    if(vis[v])  return;
    vis[v] = true;
    cnt[v/V]++;
    for(auto i : G[v])  dfs(i);
}

int main(void){
    cin >> N;
    for(int i=0;i<N;i++)    cin >> x[i] >> y[i];
    
    for(int i=0;i<N;i++){
        y[i] += V;
        G[ x[i] ].push_back(y[i]);
        G[ y[i] ].push_back(x[i]);
    }
    
    long long ans = 0;
    for(int i=0;i<V*2;i++){
        if(vis[i])  continue;
        cnt = vector<int>(2);
        dfs(i);
        ans += (long long)cnt[0] * cnt[1];
    }
    
    ans -= N;
    cout << ans << endl;
}