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
from collections import defaultdict
n = int(input())
dx = defaultdict(list)
dy = defaultdict(list)
pts = set()
for _ in range(n):
    x, y = map(int, input().split())
    dx[x].append(y)
    dy[y].append(x)
    pts.add((x, y))

vx, vy = set(), set()
cnt = 0
for x, y in pts:
    if x in vx or y in vy: continue
    sx, sy = [x], [y]
    xs, ys = set(), set()
    while sx or sy:
        while sx:
            x = sx.pop()
            if x not in xs:
                xs.add(x)
                for y in dx[x]:
                    if y not in ys:
                        sy.append(y)
        while sy:
            y = sy.pop()
            if y not in ys:
                ys.add(y)
                for x in dy[y]:
                    if x not in xs:
                        sx.append(x)
    vx.update(xs)
    vy.update(ys)
    cnt += len(xs) * len(ys)
print(cnt - len(pts))