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>
#define rep(i, n) for(int i=0; i<(n); i++)
#define int long long
#define double long double
#define P pair<long long,long long>
#define all(a) a.begin(),a.end()
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
using namespace std;

int n;
const int V=100010;
vector<int> a[V*2];
vector<int> cnt;
bool used[V*2];

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

signed main(void){
    cin.tie(0); cout.tie(0);
    ios::sync_with_stdio(false);

    cin>>n;
    rep(i,n){
        int x,y; cin>>x>>y;
        a[x].push_back(y+V);
        a[y+V].push_back(x);
    }

    int ans=0;
    rep(i,V*2){
        cnt=vector<int>(2);
        dfs(i);
        ans+=cnt[0]*cnt[1];
    }

    cout<<ans-n<<endl;
    return 0;

}