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;

const int N = 200 * 1000 + 17;
const int MOD = 1000 * 1000 * 1000 + 7;

int n = 100000, m;
vector<int> g[N];
int cnt[2];
bool used[N];

bool read() {
    if (!(cin >> m))
        return false;

    for (int i = 0; i < m; ++i) {
        int u, v;
        cin >> u >> v;
        g[u].emplace_back(n + v);
        g[n + v].emplace_back(u);
    }

    return true;
}

void dfs(int v) {
    used[v] = true;
    ++cnt[(v - 1) / n];

    for (auto to : g[v])
        if (!used[to])
            dfs(to);
}

void solve() {
    auto ans = 0ll;

    for (int x = 1; x <= n; ++x)
        if (!used[x]) {
            cnt[0] = cnt[1] = 0;
            dfs(x);
            ans += 1ll * cnt[0] * cnt[1];
        }
    
    cout << ans - m << endl;
}

int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    while (read())
        solve();

    return 0;
}

