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
import copy

N=int(input())
P=[list(map(int,input().split()))+[i] for i in range(N)]
Q=copy.deepcopy(P)

#UnionFind

Group=[i for i in range(N)]#グループ分け.Group[i]=jのときiとjは同じグループ

def find(x):#find(a)=find(b)のとき同じグループ
    while Group[x] != x:
        x=Group[x]
    return x

def Union(x,y):  # xとyが同じグループになるよう更新
    if find(x) != find(y):
        Group[find(y)]=Group[find(x)]=min(find(y),find(x))

P.sort(key=lambda x:x[0])

for i in range(1,N):
    if P[i][0]==P[i-1][0]:
        Union(P[i][2],P[i-1][2])


P.sort(key=lambda x:x[1])

for i in range(1,N):
    if P[i][1]==P[i-1][1]:
        Union(P[i][2],P[i-1][2])

#print([find(i) for i in range(N)])
#F=[find(i) for i in range(N)]
#S=set(F)

B=[[] for i in range(N)]

for i in range(N):
    B[find(i)].append(i)

ANS=0
for BEL in B:
    X=[]
    Y=[]
    for b in BEL:
        X.append(Q[b][0])
        Y.append(Q[b][1])

    #print(BEL,X,Y)
    ANS+=len(set(X))*len(set(Y))-len(BEL)
print(ANS)
        
