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;
using ll = long long;
#define rep(i, n) for (int i = 0; i < (n); i++)

vector<int> g[200001];
bool visited[200001];
ll n, r, c, p;

void dfs(int u) {
    if (visited[u]) return;
    visited[u] = true;
    if (u <= 100000)
        ++r;
    else
        ++c;
    for (int v : g[u]) {
        ++p;
        dfs(v);
    }
}

int main() {
    cin >> n;
    for (int i = 0; i < n; ++i) {
        int x, y;
        cin >> x >> y;
        g[x].push_back(y + 100000);
        g[y + 100000].push_back(x);
    }
    ll ans = 0;
    for (int u = 1; u <= 200000; ++u) {
        r = c = p = 0;
        dfs(u);
        ans += r * c - p / 2;
    }
    cout << ans << endl;
}