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
n = int(input())

pos_max = 100000

dic = dict()

for _ in range(n):
    x, y = map(int, input().split())
    y += pos_max
    if not x in dic:
        dic[x] = set()
    if not y in dic:
        dic[y] = set()
    dic[x].add(y)
    dic[y].add(x)
checked = set()

count = 0
for x in dic:
    if x in checked:
        continue
    checked.add(x)

    member = {x}
    next_items = {x}
    while next_items:
        v = next_items.pop()
        member |= dic[v]
        next_items |= dic[v] - checked
        checked |= dic[v]
    x_n = sum(w <= pos_max for w in member)
    y_n = sum(w >  pos_max for w in member)
    count += x_n * y_n
print(count - n)