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


def dfs(s, links, dots, checked):
    q = [s]
    xs = set()
    ys = set()
    cnt = 0
    while q:
        v = q.pop()
        if checked[v]:
            continue
        checked[v] = True

        x, y = dots[v]
        xs.add(x)
        ys.add(y)
        cnt += 1
        q.extend(links[v])
    return max(0, len(xs) * len(ys) - cnt)


n = int(input())
dots = [tuple(map(int, line.split())) for line in sys.stdin]
xs = {}
ys = {}
links = [set() for _ in [0] * n]
for i, (x, y) in enumerate(dots):
    if x in xs:
        links[i].add(xs[x])
        links[xs[x]].add(i)
    else:
        xs[x] = i
    if y in ys:
        links[i].add(ys[y])
        links[ys[y]].add(i)
    else:
        ys[y] = i

checked = [False] * n
ans = 0
for i in range(n):
    if checked[i]:
        continue
    ans += dfs(i, links, dots, checked)
print(ans)
