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;
typedef long long ll;
typedef pair<ll,ll> pll;
const int mn=2e5+10;
vector<int>ix[mn],iy[mn];
int x[mn],y[mn];
bool vis[mn];
ll tot;
set<int>sx,sy;
void dfs(int n){
    vis[n]=1;
    sx.insert(x[n]);
    sy.insert(y[n]);
    for(int k:ix[x[n]])if(!vis[k])dfs(k);
    for(int k:iy[y[n]])if(!vis[k])dfs(k);
}
int main(){
    int n,i;
    scanf("%d",&n);
    for(i=0;i<n;i++){
        scanf("%d%d",x+i,y+i);
        ix[x[i]].push_back(i);
        iy[y[i]].push_back(i);
    }
    for(i=0;i<n;i++){
        if(!vis[i]){
            dfs(i);
            tot+=sx.size()*sy.size();
            sx.clear();
            sy.clear();
        }
    }
    printf("%lld",tot-n);
}