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
import sys
sys.setrecursionlimit(10**6)

n = int(input())
edges = [[] for i in range(2*10**5)]
for i in range(n):
    xi, yi = map(int, input().split())
    xi -= 1
    yi = yi - 1 + 10**5
    edges[xi].append(yi)
    edges[yi].append(xi)

already = set([])
def dfs(x, y, node):
    if node in already:
        return x, y
    if node < 10**5:
        x += 1
    else:
        y += 1
    already.add(node)
    for next_node in edges[node]:
        x, y = dfs(x, y, next_node)
    return x, y

num = 0
for i in range(10**5):
    x, y = dfs(0, 0, i)
    num += x * y
print(num-n)