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
// Author: πα

#include <bits/stdc++.h>
using namespace std;
#define long int64_t

const int N = 1e5 + 5;
vector<int> g[N+N];
bool vis[N+N];
int cl, cr, ce;

void dfs(int u)
{
    if(vis[u])
        return;
    vis[u] = 1;
    if(u < N)
        cl++, ce += g[u].size();
    else
        cr++;
    for(auto v : g[u])
        dfs(v);
}

int main()
{
    int n, u, v, i;
    long ans = 0;
    cin >> n;
    for(i = 0; i < n; ++i)
    {
        cin >> u >> v;
        v += N;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    for(i = 1; i < N; ++i)
        if(not vis[i])
        {
            cl = cr = ce = 0;
            dfs(i);
            ans += 1LL * cl * cr - ce;
        }
    cout << ans;
    return 0;
}
