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>
using namespace std;
constexpr int ybit = 128 << 10;

auto connected_from = vector<vector<int>>(2 * ybit);
auto visited = vector<char>(2 * ybit);

void traverse(int v, int &nx, int &ny) {
    if (visited[v])
        return;
    visited[v] = true;
    (v & ybit ? ny : nx)++;

    for (auto u : connected_from[v])
        traverse(u, nx, ny);
}

int main() {
    int n;
    cin >> n;

    for (int i = 0; i < n; ++i) {
        int x, y;
        cin >> x >> y;

        y |= ybit;

        connected_from[x].push_back(y);
        connected_from[y].push_back(x);
    }

    auto cnt = 0LL;
    for (auto v = 0; v < 2 * ybit; ++v) {
        if (visited[v])
            continue;
        int nx = 0, ny = 0;
        traverse(v, nx, ny);
        cnt += 1LL * nx * ny;
    }

    cout << (cnt - n) << "\n";
    return 0;
}
