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 UnionFindTree:
    """Disjoint-Set Data Structure
       Union-Find Tree
       complexity:
          init: O(n)
          find, unite, same: O(alpha(n))
       used in SRM505 div.2 900, ATC001 A, DSL1A(AOJ)
    """
    def __init__(self, n):
        self.par = list(range(n))  # parent
        self.rank = [0] * n  # depth of tree

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

    def unite(self, x, y):
        x, y = self.find(x), self.find(y)
        if x == y:
            return
        if self.rank[x] < self.rank[y]:
            self.par[x] = y
        else:
            self.par[y] = x
            if self.rank[x] == self.rank[y]:
                self.rank[x] += 1

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


N = int(input())
M = 10 ** 5
uft = UnionFindTree(N + 2 * M)

for i in range(N):
    x, y = map(int, input().split())
    x, y = x-1, y-1
    uft.unite(i, x + N)
    uft.unite(i, y + N + M)
x_dict, y_dict = dict(), dict()
for x in range(N, M + N):
    par_x = uft.find(x)
    x_dict[par_x] = x_dict.get(par_x, 0) + 1
for y in range(N + M, N + 2 * M):
    par_y = uft.find(y)
    y_dict[par_y] = y_dict.get(par_y, 0) + 1
ans = -N
for r in x_dict.keys():
    ans += x_dict.get(r, 0) * y_dict.get(r, 0)
print(ans)
