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():
    n=1
    par=[0]
    rnk=[0]
    def __init__(self,size):
        self.n=size
        self.par=[i for i in range(self.n)]
        self.rnk=[0 for i in range(self.n)]
    def find(self,vertex1):
        if self.par[vertex1]==vertex1:
            return vertex1
        else:
            self.par[vertex1]=self.find(self.par[vertex1])
            return self.par[vertex1]
    def unite(self,vertex1,vertex2):
        vertex1=self.find(vertex1)
        vertex2=self.find(vertex2)
        if vertex1==vertex2:
            return
        if (self.rnk[vertex1]<self.rnk[vertex2]):
            self.par[vertex1]=vertex2
        else:
            self.par[vertex2]=vertex1
            if(self.rnk[vertex1]==self.rnk[vertex2]):
                self.rnk[vertex1]+=1
    def same(self,vetrex1,vertex2):
        return self.find(vetrex1)==self.find(vertex2)
N=int(input())
P=[(i,)+tuple(map(int,input().split())) for i in range(N)]
G=UnionFind(N)
P.sort(key=lambda x:x[1])
for i in range(N-1):
    if P[i][1]==P[i+1][1]:
        G.unite(P[i][0],P[i+1][0])
P.sort(key=lambda x:x[2])
for i in range(N-1):
    if P[i][2]==P[i+1][2]:
        G.unite(P[i][0],P[i+1][0])
P.sort(key=lambda x:x[0])
Dx=dict()
Dy=dict()
for i in range(N):
    j=G.find(i)
    if j in Dx:
        Dx[j].add(P[i][1])
    else:
        Dx[j]={P[i][1]}
    if j in Dy:
        Dy[j].add(P[i][2])
    else:
        Dy[j]={P[i][2]}
ans=-N
for j in Dx:
    ans+=len(Dx[j])*len(Dy[j])
print(ans)

    