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


class UnionFind():
    n = 1
    par = [0]
    rnk = [0]

    def __init__(self, size):
        self.n = size
        self.par = [i for i in range(self.n)]
        self.rnk = [0 for i in range(self.n)]

    def find(self, vertex1):
        if self.par[vertex1] == vertex1:
            return vertex1
        else:
            self.par[vertex1] = self.find(self.par[vertex1])
            return self.par[vertex1]

    def unite(self, vertex1, vertex2):
        vertex1 = self.find(vertex1)
        vertex2 = self.find(vertex2)
        if vertex1 == vertex2:
            return
        if (self.rnk[vertex1] < self.rnk[vertex2]):
            self.par[vertex1] = vertex2
        else:
            self.par[vertex2] = vertex1
            if (self.rnk[vertex1] == self.rnk[vertex2]):
                self.rnk[vertex1] += 1

    def same(self, vetrex1, vertex2):
        return self.find(vetrex1) == self.find(vertex2)


N = int(input())
P = [(i,) + tuple(map(int, input().split())) for i in range(N)]
G = UnionFind(N)
P.sort(key=lambda x: x[1])
for i in range(N - 1):
    if P[i][1] == P[i + 1][1]:
        G.unite(P[i][0], P[i + 1][0])
P.sort(key=lambda x: x[2])
for i in range(N - 1):
    if P[i][2] == P[i + 1][2]:
        G.unite(P[i][0], P[i + 1][0])
P.sort(key=lambda x: x[0])
Dx = defaultdict(set)
Dy = defaultdict(set)
for i in range(N):
    j = G.find(i)
    Dx[j].add(P[i][1])
    Dy[j].add(P[i][2])
ans = -N
for j in Dx:
    ans += len(Dx[j]) * len(Dy[j])
print(ans)