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;

vector<int> graf[200005];

int zliczanie[2], krawedzie, odwiedzony[200005];

void dfs(int x){
    zliczanie[odwiedzony[x] % 2]++;
    for (int i : graf[x]){
        if (!odwiedzony[i]){
            odwiedzony[i] = odwiedzony[x] + 1;
            dfs(i);
        }
        krawedzie++;
    }
}

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n, a, b;
    cin >> n;
    for (int i = 0; i < n; i++){
        cin >> a >> b;
        graf[a].push_back(b + 100000);
        graf[b + 100000].push_back(a);
    }
    long long wynik = 0;
    for (int i = 1; i <= 200000; i++)
        if (!odwiedzony[i]){
            zliczanie[0] = 0;
            zliczanie[1] = 0;
            krawedzie = 0;
            odwiedzony[i] = 1;
            dfs(i);
            krawedzie /= 2;
            wynik += (long long)zliczanie[0] * zliczanie[1] - krawedzie;
        }
    cout << wynik << "\n";
    return 0;
}