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.p = [i for i in range(n)]
        self.components = n  # 連結成分の数

    def root(self, x):
        if self.p[x] == x:
            return x
        else:
            self.p[x] = self.root(self.p[x])
            return self.p[x]
    
    def unite(self, x, y):
        x, y = self.root(x), self.root(y)
        if x != y:
            self.p[x] = y
            self.components -= 1
    
    def same(self, x,y):
        return (self.root(x) == self.root(y))

N = int(input())
MAX = 100010
UF = UnionFind(2*MAX)

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


cnt_x, cnt_y = [0]*(2*MAX), [0]*(2*MAX)

for i in range(MAX):
    cnt_x[UF.root(i)] += 1

for i in range(MAX,2*MAX):
    cnt_y[UF.root(i)] += 1

print(sum([cnt_x[i]*cnt_y[i] for i in range(2*MAX)]) - N)