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;
#define ll long long
#define db double
template<class T> void chmin(T& x, T y){if(x > y) x = y;}
template<class T> void chmax(T& x, T y){if(x < y) x = y;}

int N, x[100010], y[100010];
vector<int> s[100010], t[100010];
bool used[100010];

void dfs(int idx, set<int>& xs, set<int>& ys){
    if(used[idx]) return;
    used[idx] = true;
    xs.insert(x[idx]);
    ys.insert(y[idx]);

    for(auto u : s[x[idx]]){
        dfs(u, xs, ys);
    }

    for(auto u : t[y[idx]]){
        dfs(u, xs, ys);
    }
}

int main(){
    cin >> N;
    for(int i = 0; i < N; i++){
        cin >> x[i] >> y[i];
        s[x[i]].push_back(i);
        t[y[i]].push_back(i);
    }

    ll ans = 0;
    for(int i = 0; i < N; i++){
        if(used[i]) continue;
        set<int> xs, ys;
        dfs(i, xs, ys);
        ll a = xs.size(), b = ys.size();
        
        ans += a * b;
    }
    cout << ans - N << endl;
    return 0;
}