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>
#define rep(i,n) for(int i = 0; i <  (n); i++)
using namespace std;
using ll = long long;

const int V=100005;
vector<int> to[2*V];
bool visited[2*V];
vector<int> cnt(2);

void dfs(int i)
{
    if (visited[i]) return;
    visited[i] = true;
    cnt[i/V]++;
    for (auto next: to[i]) dfs(next);
}

int main()
{
    int n;
    cin >> n;
    rep(i,n)
    {
        int x, y;
        cin >> x >> y;
        y += V;
        to[x].push_back(y);
        to[y].push_back(x);
    }
    ll ans = 0;
    rep(i, V*2)
    {
        if (visited[i]) continue;
        cnt = {0, 0};
        dfs(i);
        ans += (ll) cnt[0] * cnt[1];
    }
    ans -= n;
    cout << ans << endl;
}