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
N = int(input())
XY = [tuple(map(int,input().split())) for i in range(N)]
V = 10**5+5

class UnionFind:
    def __init__(self,N):
        self.parent = [i for i in range(N)]
        self.rank = [0] * N
        self.size_x = [1 - i//V for i in range(N)]
        self.size_y = [i//V for i in range(N)]
    def root(self,a):
        if self.parent[a] == a:
            return a
        else:
            self.parent[a] = self.root(self.parent[a])
            return self.parent[a]
    def is_same(self,a,b):
        return self.root(a) == self.root(b)
    def unite(self,a,b):
        ra = self.root(a)
        rb = self.root(b)
        if ra == rb: return
        if self.rank[ra] < self.rank[rb]:
            self.parent[ra] = rb
            self.size_x[rb] += self.size_x[ra]
            self.size_y[rb] += self.size_y[ra]
        else:
            self.parent[rb] = ra
            if self.rank[ra] == self.rank[rb]: self.rank[ra] += 1
            self.size_x[ra] += self.size_x[rb]
            self.size_y[ra] += self.size_y[rb]

uf = UnionFind(V*2)

for x,y in XY:
    uf.unite(x,y+V)
for i in range(V*2):
    uf.root(i)

ans = 0
used = set()
for i in range(V*2):
    r = uf.root(i)
    if r in used: continue
    used.add(r)
    ans += uf.size_x[r] * uf.size_y[r]
ans -= N
print(ans)