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 Union_Find():
    def __init__(self, num):
        self.par = [-1]*(num+1)
        self.siz = [1]*(num+1)

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

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

    def union(self, x, y):
        rx = self.find(x)
        ry = self.find(y)
        if rx != ry:
            if self.par[rx] < self.par[ry]:
                self.par[ry] = rx
                self.siz[rx] += self.siz[ry]
            elif self.par[rx] > self.par[ry]:
                self.par[rx] = ry
                self.siz[ry] += self.siz[rx]
            else:
                self.par[rx] -= 1
                self.par[ry] = rx
                self.siz[rx] += self.siz[ry]
        return

    def size(self, x):
        return self.siz[self.find(x)]

n = int(input())
uf = Union_Find(n)

d = [list(map(int, input().split())) for i in range(n)]
x = [[] for i in range(100001)]
y = [[] for i in range(100001)]

for i in range(n):
    xi, yi = d[i]
    x[xi].append(i+1)
    y[yi].append(i+1)


for i in range(1, 100001):
    for j in range(len(x[i])-1):
        uf.union(x[i][j], x[i][j+1])

for i in range(1, 100001):
    for j in range(len(y[i])-1):
        uf.union(y[i][j], y[i][j+1])

par = [[] for i in range(n+1)]
for i in range(1, n+1):
    par[uf.find(i)].append(d[i-1])

ans = -n
for i in range(1, n+1):
    x = set()
    y = set()
    for xi, yi in par[i]:
        x.add(xi)
        y.add(yi)

    ans += len(x)*len(y)

print(ans)
