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>
#include<iomanip>
#include<numeric>

using namespace std;
using ll = long long;
constexpr ll mod = 1e9+7;
constexpr ll md = mod;
constexpr ll inf = 1e15;
vector<int> E[200200];
bool visited[200200];
vector<int> c;
void dfs(int v){
    visited[v] = true;
    for(int u : E[v]){
            if(!visited[u])dfs(u);
    }
    c[v/100050]++;
}
int main(){
    int n;
    cin>>n;
    for(int i=0;i<n;++i){
        int x,y;
        cin>>x>>y;
        y+=100050;
        E[x].push_back(y);
        E[y].push_back(x);
    }
    ll ret = 0;
    for(int i=1;i<=200100;++i){
        if(!visited[i]){
            c = {0,0};
            dfs(i);
            ret += 1ll*c[0]*c[1];
        }
    }
    cout << ret - n << endl;
    return 0;
}

