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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;

#define REP(i, n) for(int i = 0; i < (int)(n); ++i)
#define FOR(i, m, n) for(int i = (int)(m); i < (int)(n); ++i)

const int max_v = 1e+5;
vector<bool> visited(2*max_v+1, false);

void dfs(int pos, vector<long long>& num_v, vector<vector<int>>& G){
    if(visited[pos]) return;
    visited[pos] = true;
    ++num_v[(pos-1)/max_v];
    for(int v: G[pos]) dfs(v, num_v, G);
}

int main()
{
    int N;
    cin >> N;
    vector<vector<int>> G(2*max_v+1);  // y は +max_v して頂点にする
    REP(i, N){
        int x, y;
        cin >> x >> y;
        y += max_v;
        G[x].push_back(y);
        G[y].push_back(x);
    }

    long long ans = 0;
    // 答えはグラフを連結成分ごとに完全二部グラフにするのに必要な本数の和
    // つまり、連結成分ごとの x, y の頂点数が必要
    REP(i, 2*max_v+1){
        vector<long long> num_v(2);
        dfs(i, num_v, G);
        ans += num_v[0]*num_v[1];
    }
    cout << ans-N << endl;
    
    return 0;
}