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.n = n
        self.p = [e for e in range(n)]
        self.rank = [0] * n
        self.size = [1] * n

    def same(self, u, v):
        return self.find_set(u) == self.find_set(v)

    def unite(self, u, v):
        u = self.find_set(u)
        v = self.find_set(v)

        if u == v:
            return

        if self.rank[u] > self.rank[v]:
            self.p[v] = u
            self.size[u] += self.size[v]
        else:
            self.p[u] = v
            self.size[v] += self.size[u]
            if self.rank[u] == self.rank[v]:
                self.rank[v] += 1

    def find_set(self, u):
        if u != self.p[u]:
            self.p[u] = self.find_set(self.p[u])

        return self.p[u]

    def update_p(self):
        for u in range(self.n):
            self.find_set(u)

    def get_size(self, u):
        return self.size[self.find_set(u)]


n = int(input())
xy = [list(map(int, input().split())) for _ in range(n)]
MAX = 10 ** 5

uf = UnionFind(MAX * 2)

for x, y in xy:
    x -= 1
    y -= 1
    uf.unite(x, y + MAX)

x_size = [0] * MAX * 2
y_size = [0] * MAX * 2
for i in range(MAX * 2):
    root = uf.find_set(i)
    if i < MAX:
        x_size[root] += 1
    else:
        y_size[root] += 1

ans = -n
for i in range(MAX * 2):
    ans += x_size[i] * y_size[i]

print(ans)
