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())
MAX = 10**5
adj = [[[] for u in range(MAX+1)] for t in range(2)]
vis = [[False for u in range(MAX+1)] for t in range(2)]
for i in range(n):
	x, y = map(int,input().split())
	adj[0][x].append(y)
	adj[1][y].append(x)

import sys
sys.setrecursionlimit(4*10**5+1)

def dfs(t,u): # Left, Right, Edges
	vis[t][u] = True

	ct = [0,0,0]
	ct[t] += 1
	for v in adj[t][u]:
		ct[2] += 1
		if not vis[t^1][v]:
			ctv = dfs(t^1,v)

			ct[0] += ctv[0]
			ct[1] += ctv[1]
			ct[2] += ctv[2]

	return ct

ans = 0
for u in range(1,MAX+1):
	if not vis[0][u]:
		L, R, M = dfs(0,u)
		ans += L*R-M//2

print(ans)