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 <iostream>
#include <cstdio>
#include <vector>

using namespace std;

typedef long long ll;

const ll M = 100000;
ll n;
vector<ll> ve[200005];
ll ans;
bool v[200005];
ll x, y;

void dfs(ll p)
{
    if(v[p]) return;
    v[p] = true;
    if(p > M) y++;
    else x++;
    for(ll i = 0; i < ve[p].size(); i++)
    {
        dfs(ve[p][i]);
    }
}

int main()
{
    scanf("%lld", &n);
    for(ll i = 0; i < n; i++)
    {
        ll t1, t2;
        scanf("%lld%lld", &t1, &t2);
        t2 += M;
        ve[t1].push_back(t2);
        ve[t2].push_back(t1);
    }
    for(ll i = 1; i <= 2 * M; i++)
    {
        if(!v[i]) x = y = 0, dfs(i), ans += x * y;
    }
    cout<<ans - n<<endl;
    return 0;
}
