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
class UnionFind:
    def __init__(self, n):
        self.ps = [-1] * (n + 1)

    def find(self, x):
        if self.ps[x] < 0:
            return x
        else:
            self.ps[x] = self.find(self.ps[x])
            return self.ps[x]

    def unite(self, x, y):
        x = self.find(x)
        y = self.find(y)
        if x == y:
            return False
        if self.ps[x] > self.ps[y]:
            x, y = y, x
        self.ps[x] += self.ps[y]
        self.ps[y] = x
        return True

    def same(self, x, y):
        return self.find(x) == self.find(y)

    def size(self, x):
        x = self.find(x)
        return -self.ps[x]

n = int(input())
v = 10**5 + 5
uf = UnionFind(v*2 + 10)
for _ in range(n):
  x, y = map(int, input().split())
  uf.unite(x, y+v)
d = dict()
for i in range(v*2):
  if uf.size(i) > 1:
    p = uf.find(i)
    if p not in d:
      d[p] = [0, 0]
    d[p][i > v] += 1
print(sum(x*y for x,y in d.values()) - n)