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: int) -> None:
        #マイナスならルートでその深さを表す。プラスならルートのノードを指す。
        self.nodes = [-1]*n
 
    def find_root(self, x: int) -> int:
        if self.nodes[x] < 0:
            return x
        else:
            self.nodes[x] = self.find_root(self.nodes[x])
            return self.nodes[x]
 
    def unite(self, x: int, y: int) -> None:
        x, y = self.find_root(x), self.find_root(y)
        #浅い方につなぐ
        if (x != y):
            if (x < y):
                self.nodes[y] += self.nodes[x]
                self.nodes[x] = y
            else:
                self.nodes[x] += self.nodes[y]
                self.nodes[y] = x
                
    def get_num(self, x: int) -> int:
        return -(self.nodes[x])

MAX = 10**5+1

N = int(input())
uf = UnionFind(2*MAX)

for i in range(N):
  x,y = map(int, input().split())
  uf.unite(x,y+MAX)

#数え上げ
num_x = [0]*2*MAX
num_y = [0]*2*MAX
for i in range(MAX):
  num_x[uf.find_root(i)] += 1
  num_y[uf.find_root(i+MAX)] += 1
  #print(num_x[uf.find_root(i)], num_y[uf.find_root(i+MAX)])

ans = 0
for i in range(2*MAX):
  ans += num_x[i] * num_y[i]

print(ans - N)
