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<vector>
#include<map>
#include<set>
#include<string>
#include<utility>
#include<algorithm>
#include<cstdio>
#include<iomanip>
#include<queue>
#include<stack>

#define ll int64_t
#define Rep(i, n) for (ll i = 0; i < n; i++)

using namespace std;

ll v = 1e5+10;
vector<ll> cnt(2, 0);
vector<vector<ll>> Graph(v*2);
vector<bool> visited(v*2, false);

void dfs (ll V) {
    if (visited[V]) return;
    visited[V] = true;
    cnt[V/v]++;
    for (auto e : Graph[V]) {
        dfs(e);
    }
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    ll N;
    cin >> N;
    Rep (i, N) {
        ll x, y;
        cin >> x >> y;
        y += v;
        Graph[x].push_back(y);
        Graph[y].push_back(x);
    }

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

    // cout << ans << "\n";

    cout << ans - N << "\n";
}