Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3
1 1
5 1
5 5

Sample Output 1

1

By choosing a = 1, b = 1, c = 5, d = 5, we can add a dot at (1, 5). We cannot do the operation any more, so the maximum number of operations is 1.

Sample Input 2

2
10 10
20 20

Sample Output 2

0

There are only two dots, so we cannot do the operation at all.

Sample Input 3

9
1 1
2 1
3 1
4 1
5 1
1 2
1 3
1 4
1 5

Sample Output 3

16

We can do the operation for all choices of the form a = 1, b = 1, c = i, d = j (2 \leq i,j \leq 5), and no more. Thus, the maximum number of operations is 16.

Solution in PyPy3 (2.4.0) (judge verdict: Wrong Answer)

N = int(input())
Points = [list(map(int, input().split())) for i in range(N)]

X_dict = {i: [] for i in range(1, 10 ** 5 + 1)}
Y_dict = {i: [] for i in range(1, 10 ** 5 + 1)}
for i, (x, y) in enumerate(Points):
    X_dict[x].append((y, i))
    Y_dict[y].append((x, i))


ans = 0
Visited = [False] * N
for i, (x, y) in enumerate(Points):
    if Visited[i]:
        continue

    stack = [('x', x), ('y', y)]
    cnt = 1
    x_set, y_set = {x}, {y}
    Visited[i] = True
    while stack:
        s, v = stack.pop()
        if s == 'x':
            for yd, j in Y_dict[v]:
                if Visited[j]:
                    continue
                else:
                    cnt += 1
                    y_set.add(yd)
                    stack.append(('y', yd))
                    Visited[j] = True
        else:
            for xd, j in X_dict[v]:
                if Visited[j]:
                    continue
                else:
                    cnt += 1
                    x_set.add(xd)
                    stack.append(('x', xd))
                    Visited[j] = True

    ans += len(x_set) * len(y_set) - cnt

print(ans)