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)]

class UnionFind:
    def __init__(self,N):
        self.parent = [i for i in range(N)]
        self._size = [1] * N
        self.count = 0
    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._size[ra] < self._size[rb]: ra,rb = rb,ra
        self._size[ra] += self._size[rb]
        self.parent[rb] = ra
        self.count += 1
    def size(self,a):
        return self._size[self.root(a)]

M = 10**5
uf = UnionFind(2*M)
for x,y in XY:
    a,b = x-1,y-1+M
    if uf.is_same(a,b): continue
    uf.unite(a,b)
for i in range(2*M):
    uf.root(i)

rt = []
for i in range(2*M):
    rt.append(uf.root(i))
from collections import Counter
xr = Counter(rt[:M])
yr = Counter(rt[M:])

ans = 0
for k,v in xr.items():
    ans += yr[k] * v
ans -= N
print(ans)