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 operator import itemgetter

class UnionFindTree():
    def __init__(self, N):
        self.__parent_of = [None] * N
        self.__rank_of = [0] * N
        self.__size = [1] * N

    def root(self, value):
        if self.__parent_of[value] is None:
            return value
        else:
            self.__parent_of[value] = self.root(self.__parent_of[value])
            return self.__parent_of[value]

    def unite(self, a, b):
        r1 = self.root(a)
        r2 = self.root(b)
        if r1 != r2:
            if self.__rank_of[r1] < self.__rank_of[r2]:
                self.__parent_of[r1] = r2
                self.__size[r2] += self.__size[r1]
            else:
                self.__parent_of[r2] = r1
                self.__size[r1] += self.__size[r2]
                if self.__rank_of[r1] == self.__rank_of[r2]:
                    self.__rank_of[r1] += 1

    def is_same(self, a, b):
        return self.root(a) == self.root(b)

    def size(self, a):
        return self.__size[self.root(a)]

    def groups(self):
        groups = {}
        for k in range(len(self.__parent_of)):
            r = self.root(k)
            if r not in groups:
                groups[r] = []
            groups[r].append(k)
        return [groups[x] for x in groups]


def main():
    N = int(input())
    uft = UnionFindTree(N)
    P = [None] * N
    for i in range(N):
        x, y = map(int, input().split())
        P[i] = (i, x, y)
    Q = sorted(P, key=itemgetter(1))
    for i in range(1, N):
        if Q[i-1][1] == Q[i][1]:
            uft.unite(Q[i-1][0], Q[i][0])
    R = sorted(P, key=itemgetter(2))
    for i in range(1, N):
        if R[i-1][2] == R[i][2]:
            uft.unite(R[i-1][0], R[i][0])
    groups = uft.groups()
    ans = 0
    for g in groups:
        if len(g) < 3:
            continue
        xs = set()
        ys = set()
        for i in g:
            _, x, y = P[i]
            xs.add(x)
            ys.add(y)
        ans += len(xs) * len(ys) - len(g)
    print(ans)


if __name__ == "__main__":
    main()
