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())
p = [list(map(int, input().split())) for _ in range(n)]
adj = [[] for _ in range(200000)]
for x, y in p:
	adj[x-1].append(y+99999)
	adj[y+99999].append(x-1)

visited = [False for _ in range(200000)]

def dfs(orig):
	stack = [orig]
	edges = len(adj[orig])
	if orig < 100000:
		x_num, y_num = 1, 0
	else:
		x_num, y_num = 0, 1

	while stack:
		v = stack.pop()
		for w in adj[v]:
			if visited[w] == False:
				visited[w] = True
				if w < 100000:
					x_num += 1
				else:
					y_num += 1
				edges += len(adj[w])
				stack.append(w)
	return x_num * y_num - (edges // 2)

ans = 0
for i in range(200000):
	if visited[i] == False and adj[i]:
		visited[i] = True
		ans += dfs(i)
print(ans)