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 vvi vector<vector<int>>
#define V 100010
#define ll long long int

vvi to(V*2);
vector<bool> visited(V*2);
vector<int> cnt(2);

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

int main() {

    int N;
    cin >> N;

    int x, y;
    for (int i = 0; i < N; i++) {
        cin >> x >> y;
        y += V;
        to[x].push_back(y);
        to[y].push_back(x);
    }

    ll ans = 0;
    for (int i = 0; i < V * 2; i++) {
        if (visited[i]) continue;
        cnt = vector<int>(2);
        dfs(i);
        ans += (ll)cnt[0]*cnt[1];
    }

    cout << ans - N << endl;

    return 0;
}
