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())

m = 10**5

edge = [[] for i in range(2*m)]

for i in range(n):
    x,y = map(int,input().split())
    x -= 1
    y -= 1 
    y += m
    edge[x].append(y)
    edge[y].append(x)

used = [False]*2*m

res = 0

for e in range(m):
    if used[e]:
        continue
    used[e] = True
    tank = []
    new = [e]

    px = 0
    py = 0

    if e >= 10**5:
        py += 1
    else:
        px += 1

    while len(new) > 0:
        tank = []
        for fr in new:
            for go in edge[fr]:
                if not used[go]:
                    tank.append(go)
                    used[go] = True
                    if go >= 10**5:
                        py += 1
                    else:
                        px += 1

        new = tank
    
    res += px*py
print(res-n)
