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 bisect import bisect_left
N = int(input())
points = list(tuple(map(int,input().split())) for _ in range(N))

X = sorted(set(x for x,y in points))
Y = sorted(set(y for x,y in points))

points = set((bisect_left(X, x), bisect_left(Y, y)) for x,y in points)

xty = [list() for _ in range(len(X))]
ytx = [list() for _ in range(len(Y))]
for x,y in points:
    xty[x].append(y)
    ytx[y].append(x)

area = 0
visited = set()

for p in points:
    if p in visited:
        continue

    stack = [p]

    dx = set()
    dy = set()

    while stack:
        x,y = stack.pop()
        visited.add((x,y))
        if x not in dx:
            for yy in xty[x]:
                if yy != y:
                    stack.append((x,yy))
            dx.add(x)
        if y not in dy:
            for xx in ytx[y]:
                if xx != x:
                    stack.append((xx,y))
            dy.add(y)
    area += len(dx)*len(dy)

print(area-len(points))