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())
class unionfind:
    def __init__(self, n):
        self.n = n
        self.root = [-1]*(n+1)
        self.rnk = [0]*(n+1)
    def find(self, x):
        if(self.root[x] < 0):
            return x
        else:
            self.root[x] = self.find(self.root[x])
            return self.root[x]
    def unite(self, x, y):
        x = self.find(x)
        y = self.find(y)
        if(x == y):
            return 
        elif(self.rnk[x] > self.rnk[y]):
            self.root[x] += self.root[y]
            self.root[y] = x
        else:
            self.root[y] += self.root[x]
            self.root[x] = y
            if(self.rnk[x] == self.rnk[y]):
                self.rnk[y] += 1
    def same(self, x, y):
        return self.find(x) == self.find(y)
    def count(self, x):
        return -self.root[self.find(x)]
    
uf=unionfind(n)
import collections
dx=collections.defaultdict(int)
dy=collections.defaultdict(int)
l=[list(map(int,input().split())) for i in range(n)]
for i,j in enumerate(l,1):
    x,y=j
    if dx[x]:
        uf.unite(dx[x],i)
    else:
        dx[x]=i
    if dy[y]:
        uf.unite(dy[y],i)
    else:
        dy[y]=i
dx=collections.defaultdict(set)
dy=collections.defaultdict(set)
for i,j in enumerate(l,1):
    x,y=j
    dx[uf.find(i)].add(x)
    dy[uf.find(i)].add(y)
ans=0
for i,j in dx.items():
    ans+=len(j)*len(dy[i])-uf.count(i)
print(ans)