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
from collections import deque
input = sys.stdin.buffer.readline
sys.setrecursionlimit(10 ** 7)

N = int(input())
U = 10 ** 5
edge = [[] for _ in range(2 * U + 1)]
for _ in range(N):
    x, y = map(int, input().split())
    y += U
    edge[x].append(y)
    edge[y].append(x)

ans = 0
seen = [False] * (2 * U + 1)
for i in range(1, 2 * U + 1):
    if not edge[i] or seen[i]:
        continue
    seen[i] = True
    node = deque([i])
    x = 0
    y = 0
    while node:
        s = node.popleft()
        if s <= U:
            x += 1
        else:
            y += 1
        for t in edge[s]:
            if seen[t]:
                continue
            seen[t] = True
            node.append(t)
    ans += x * y

ans = ans - N
print(ans)