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;

typedef long long ll;

int N;
vector<int> V[200010];
int vis[200010];
int cx, cy;

void DFS(int v)
{
    if(vis[v] != 0)return;
    if(v % 2 == 1)cy++;
    else cx++;
    vis[v] = 1;
    for(int i = 0; i < V[v].size(); i++)DFS(V[v][i]);
}

int main()
{
    cin >> N;
    for(int i = 0; i < N; i++)
    {
        int a, b;
        cin >> a >> b;
        int x = 2 * a;
        int y = 2 * b + 1;
        V[x].push_back(y);
        V[y].push_back(x);
    }

    ll cnt = 0;
    for(int i = 0; i < 200000; i++)
    {
        if(vis[i] == 0)
        {
            cx = cy = 0;
            DFS(i);
            cnt += (ll)cx * (ll)cy;
        }
    }

    cout << cnt - N << endl;
    return 0;
}
