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 <set>
#include <vector>
using namespace std;
typedef long long ll;

ll N;
vector<vector<int>> v(200010);
int vis[200010] = {};

ll dfs(int n,set<int>& sx,set<int>& sy){
    ll cnt = 0;
    vis[n] = 1;
    if(n<=100000) sx.insert(n);
    else sy.insert(n);
    for(auto x:v[n]){
        cnt++;
        if(vis[x]==0) cnt += dfs(x,sx,sy);
    }
    return cnt;
}

int main(){
    cin >> N;
    int x,y;
    for(int i=0;i<N;i++){
        cin >> x >> y;
        v[x].push_back(y+100000);
        v[y+100000].push_back(x);
    }
    ll ans = 0;
    for(int i=1;i<=100000;i++){
        if((int) v[i].size()!=0 && vis[i]==0){
            set<int> sx,sy;
            ll n = dfs(i,sx,sy)/2;
            ll cx = sx.size(),cy = sy.size();
            ans += cx*cy-n;
        }
    }
    cout << ans << endl;
}