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
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__

n = int(input())
m = 10**5
G = [[] for i in range(2*m)]
for i in range(n):
    x,y = map(int, input().split())
    x,y = x-1, y-1
    G[x].append(m+y)
    G[m+y].append(x)

from collections import deque
used = [0]*(2*m)
def dfs(i):
    q = deque([i])
    x = 0
    y = 0
    edge = 0
    while q:
        cur = q.pop()
        if used[cur]:
            continue
        used[cur] = 1
        if cur<m:
            x += 1
        else:
            y += 1
        edge += len(G[cur])
        for to in G[cur]:
            if used[to] == 0:
                q.append(to)
    return x*y-edge//2

ans = 0
for i in range(2*m):
    if used[i]:
        continue
    ans += dfs(i)
print(ans)
