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

    def find(self, x):
        if self.par[x] != x:
            self.par[x] = self.find(self.par[x]) # contruction
        return self.par[x]

    def unite(self, x, y):
        x = self.find(x)
        y = self.find(y)
        self.par[x] = y

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


n = int(input())

uf = UnionFind(n)
xs = []
ys = []
x_dict = {}
y_dict = {}
for i in range(n):
    x, y = map(int, input().split())
    xs.append(x)
    ys.append(y)

    if x in x_dict:
        uf.unite(i, x_dict[x])

    if y in y_dict:
        uf.unite(i, y_dict[y])

    x_dict[x] = uf.find(i)
    y_dict[y] = uf.find(i)

# print(uf.par)
xi = defaultdict(list)
yi = defaultdict(list)
for i in range(n):
    xi[uf.find(x_dict[xs[i]])].append(xs[i])
    yi[uf.find(y_dict[ys[i]])].append(ys[i])

ans = 0
for k in xi.keys():
    ans += len(set(xi[k])) * len(set(yi[k])) - len(xi[k])

print(ans)
