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
n = int(input())
xy = [list(map(int, input().split())) for _ in range(n)]
xtoy = {}
ytox = {}
for x, y in xy:
    if x not in xtoy:
        xtoy[x] = []
    xtoy[x].append(y)
    if y not in ytox:
        ytox[y] = []
    ytox[y].append(x)
ret = 0
while len(xtoy) > 0:
    x = next(iter(xtoy))
    xs = set()
    ys = set()
    xs.add(x)
    q = [(x, None)]
    while len(q) > 0:
        (xx, yy) = q.pop(0)
        if xx is not None:
            for y in xtoy.pop(xx):
                if y not in ys:
                    ys.add(y)
                    q.append((None, y))
        if yy is not None:
            for x in ytox.pop(yy):
                if x not in xs:
                    xs.add(x)
                    q.append((x, None))
    ret += len(xs) * len(ys)
print(ret - n)
