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
input = sys.stdin.readline
M = 10**5

n = int(input())
G = [[] for _ in range(2*M+1)]
for _ in range(n):
  x, y = map(int, input().split())
  G[x].append(y+M)
  G[y+M].append(x)
seen = [False]*(2*M+1)
def dfs(v):
  stack = [v]
  seen[v] = True
  cnt_x, cnt_y = 0, 0
  while stack:
    v = stack.pop()
    if v <= M:
      cnt_x += 1
    else:
      cnt_y += 1
    for nv in G[v]:
      if seen[nv]:
        continue
      seen[nv] = True
      stack.append(nv)
  return cnt_x*cnt_y
ans = 0
for i in range(2*M+1):
  if not seen[i]:
    ans += dfs(i)
ans -= n
print(ans)