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
def main():
    n=int(input())
    xy=[list(map(int,input().split())) for _ in [0]*n]
    ixy=[[i,x,y] for i,(x,y) in enumerate(xy)]
    class unionfind():
        #size:要素数,tree：unionfind木
        def __init__(self,size):#self,要素数
            self.size=size
            self.tree_root=list(range(self.size))
            self.tree_depth=[1]*self.size
        
        #rootを探す
        def root(self,index):
            temp_list=[]
            temp=self.tree_root[index]
            while index!=temp:
                temp_list.append(index)
                index=temp
                temp=self.tree_root[index]
            for i in temp_list:
                self.tree_root[i]=index
            return index
        
        #結合
        def unite(self,index1,index2):
            r1=self.root(index1)
            r2=self.root(index2)
            if r1!=r2:
                d1,d2=self.tree_depth[r1],self.tree_depth[r2]
                if d1<=d2:
                    self.tree_root[r1]=r2
                    self.tree_depth[r2]=max(d1+1,d2)
                else:
                    self.tree_root[r2]=r1
                    self.tree_depth[r1]=max(d2+1,d1)
    
        #同じか判定
        def same(self,index1,index2):
            r1=self.root(index1)
            r2=self.root(index2)
            return r1==r2
    
        #連結成分の個数
        def component(self):
            return len({self.root(i) for i in range(self.size)})

    u=unionfind(n)
    ixy.sort(key=lambda x:x[1])
    for i in range(n-1):
        if ixy[i][1]==ixy[i+1][1]:
            u.unite(ixy[i][0],ixy[i+1][0])
    ixy.sort(key=lambda x:x[2])
    for i in range(n-1):
        if ixy[i][2]==ixy[i+1][2]:
            u.unite(ixy[i][0],ixy[i+1][0])
    tate=[set() for _ in [0]*n]
    yoko=[set() for _ in [0]*n]
    cnt=[0]*n
    for i in range(n):
        r=u.root(i)
        tate[r].add(xy[i][0])
        yoko[r].add(xy[i][1])
        cnt[r]+=1
    print(sum([len(tate[i])*len(yoko[i])-cnt[i] for i in range(n)]))
main()