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 row(x) (x)
#define col(x) ((x) + 100005)

int cmpID;
int xcnt[200005], ycnt[200005];
int vis[200005];
vector< vector<int> > adj;
void dfs(int u){
    if(vis[u]) return;
    vis[u] = 1;
    if(u <= 100000)
        xcnt[cmpID]++;
    else
        ycnt[cmpID]++;
    for(int v:adj[u])
        dfs(v);
}
int main() {
#ifndef ONLINE_JUDGE
    //freopen("input.txt", "r", stdin);
#endif
    int n;
    scanf("%d", &n);
    adj = vector< vector<int> >(200005);
    for(int i=0; i<n; i++){
        int x, y;
        scanf("%d %d", &x, &y);
        adj[row(x)].push_back(col(y));
        adj[col(y)].push_back(row(x));
    }
    for(int i=0; i<adj.size(); i++){
        if(adj[i].size() && !vis[i]) {
            dfs(i);
            cmpID++;
        }
    }
    long long res = 0;
    for(int i=0; i<cmpID; i++)
        res += xcnt[i] * 1ll * ycnt[i];
    printf("%lld\n", res - n);
    return 0;
}