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
V=10**5+5
class UnionFind:
    def __init__(self,n):
        self.parent=[i for i in range(n)]
        self.rank=[0 for _ in range(n)]
        # use for n=2*V
        self.xSize=[1 if i<V else 0 for i in range(n)]
        self.ySize=[0 if i<V else 1 for i in range(n)]

    def find(self,x):
        if self.parent[x]==x:
            return x
        else:
            parentx=self.find(self.parent[x])
            self.parent[x]=parentx
            return parentx

    def union(self,x,y):
        rootx=self.find(x)
        rooty=self.find(y)
        if rootx==rooty:
            return
        if self.rank[rootx]>self.rank[rooty]:
            self.parent[rooty]=rootx
            self.xSize[rootx]+=self.xSize[rooty]
            self.ySize[rootx]+=self.ySize[rooty]
        else:
            self.parent[rootx]=rooty
            self.xSize[rooty]+=self.xSize[rootx]
            self.ySize[rooty]+=self.ySize[rootx]
            if self.rank[rootx]==self.rank[rooty]:
                self.rank[rooty]+=1

n=int(input())
xy=[tuple(map(int,input().split())) for _ in range(n)]
uf=UnionFind(V*2)
for xi,yi in xy:
    uf.union(xi,yi+V)

# for i in range(V*2):
#     uf.find(i)

ans=0
used=set()
for i in range(V*2):
    rooti=uf.find(i)
    if rooti not in used:
        used.add(rooti)
        ans+= uf.xSize[rooti]*uf.ySize[rooti]
print(ans-n)
