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;

unordered_map<int, bool> vis;
unordered_map<int, vector<int>> eg;

ll bfs(int st) {
    queue<int> q; q.push(st); vis[st] = true;
    int now, sz[2] = {};
    while (!q.empty()) {
        now = q.front(); q.pop(); sz[now > 0]++;
        for (auto i: eg[now]) {
            if (!vis.count(i)) q.push(i), vis[i] = true;
        }
    }
    //cout << st << ":" << sz[0] << "," << sz[1] << endl;
    return 1LL * sz[0] * sz[1];
}

int main() {
    ll ans = 0, n;
    cin >> n;
    for (int i = 0, x, y; i < n; i++) {
        cin >> x >> y;
        eg[x].push_back(-y), eg[-y].push_back(x);
    }
    for (auto i: eg) {
        if (!vis.count(i.first)) ans += bfs(i.first);
    }
    cout << ans - n << endl;
    return 0;
}
