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
import sys
input = sys.stdin.readline
inpl = lambda: list(map(int,input().split()))
from collections import defaultdict

class UnionFind:
    def __init__(self, N=None):
        if N is None or N < 1:
            self.parent = defaultdict(lambda: -1)
        else:
            self.parent = [-1]*int(N)

    def root(self, n):
        if self.parent[n] < 0:
            return n
        else:
            m = self.root(self.parent[n])
            self.parent[n] = m
            return m

    def merge(self, m, n):
        rm = self.root(m)
        rn = self.root(n)
        if rm != rn:
            if -self.parent[rm] < -self.parent[rn]:
                rm, rn = rn, rm
            self.parent[rm] += self.parent[rn]
            self.parent[rn] = rm

    def size(self, n):
        return -self.parent[self.root(n)]
    
    def connected(self, m, n):
        return self.root(m) == self.root(n)
    
    def groups(self):
        if isinstance(self.parent,list):
            return list(map(lambda x: x<0, self.parent)).count(True)
        else: # self.parent: defaultdict
            return list(map(lambda x: x<0, self.parent.values())).count(True) 

class CountUp:
    def __init__(self, start=0):
        self.index = start-1

    def __call__(self):
        self.index += 1
        return self.index

N = int(input())
uf = UnionFind()
for _ in range(N):
    x, y = inpl()
    uf.merge(2*x+1,2*y)

Ng = uf.groups()
nx = [0]*Ng
ny = [0]*Ng
gi = defaultdict(CountUp())
for k in uf.parent.keys():
    if k % 2:
        nx[gi[uf.root(k)]] += 1
    else:
        ny[gi[uf.root(k)]] += 1

ans = sum([ nx[g]*ny[g] for g in range(Ng) ]) - N
print(ans)