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
/*ABC 131 F */
#include<bits/stdc++.h>
#define rep(i,n) for(int i = 0;i < (n);i++)
typedef long long ll;
using namespace std;
// typedef pair<int,int> P;
//pair<first,secont>;
const int V = 100002;
vector<int> grp[V*2];
bool visited[V*2];
vector<int> cnt(2);

void dfs(int v){
    if (visited[v] == true)return ;
    visited[v] = true;
    cnt[v/V]++;
    for (int u : grp[v])dfs(u);
}

int main(){
    int n;
    cin >> n;
    rep(i,n) {
        int x , y ;
        cin  >> x >> y;
        y += V;
        //無向グラフを作成（有向グラフを相互に作る）
        grp[x].push_back(y);
        grp[y].push_back(x);
        // rep(i,V*2) rep(j,grp[i].size()) cout << grp[i][j] << endl;
    }
    ll ans = 0;
    rep(i,V*2){
        if (visited[i] == true) continue;
        cnt = vector<int>(2);//カウンタの初期化
        dfs(i);
        ans += (ll)cnt[0]*cnt[1];
    }
    ans -= n;//元から存在した辺を除く
    cout << ans << endl;
    return 0;

}