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.v = [-1 for _ in range(n)]

    def find(self, x):
        if self.v[x] < 0:  # (負)は根
            return x
        else:
            self.v[x] = self.find(self.v[x])
            return self.v[x]

    def unite(self, x, y):
        x = self.find(x)
        y = self.find(y)
        if x == y:
            return
        if -self.v[x] < -self.v[y]:
            x, y = y, x
        self.v[x] += self.v[y]
        self.v[y] = x

    def root(self, x):
        return self.v[x] < 0

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


mx = 10 ** 5 + 10

n = int(input())

uf = UnionFind(mx * 2)

for _ in range(n):
    x, y = map(int, input().split())
    uf.unite(x, y + mx)

vx = [0] * (mx * 2)
vy = [0] * (mx * 2)

for i in range(mx):
    root = uf.find(i)
    vx[root] += 1

for i in range(mx, mx * 2):
    root = uf.find(i)
    vy[root] += 1

ans = -n
for xx, yy in zip(vx, vy):
    ans += xx * yy
print(ans)
