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 Python (3.4.3) (judge verdict: Runtime Error)

MAXVAL = 10**5
N = int(input())
edge = [[] for _ in range(MAXVAL*2)]
visited = [0] * (MAXVAL*2)
ans = 0
xnum = 0; ynum = 0; nodes = 0

def dfs(n):
    global xnum, ynum, nodes
    if visited[n] == 1:
        return True
    if n < MAXVAL:
        xnum += 1
    else:
        ynum += 1
    visited[n] = 1
    for item in edge[n]:
        nodes += 1
        dfs(item)
    return True

for i in range(N):
    x, y = [int(item) for item in input().split()]
    x -= 1; y -= 1
    edge[x].append(y+MAXVAL)
    edge[y+MAXVAL].append(x)

for i in range(MAXVAL*2):
    if visited[i] == 1:
        continue
    xnum = 0; ynum = 0; nodes = 0
    dfs(i)
    if nodes >= 3:
        ans += xnum * ynum - nodes // 2
print(ans)