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

from collections import deque

n = int(input())
xy = [list(map(int,input().split())) for _ in range(n)]

x_list = {}
y_list = {}
for i in range(1,10**5+1):
    x_list[i] = set()
    y_list[i] = set()

for i in range(n):
    x,y = xy[i]
    x_list[x].add(i)
    y_list[y].add(i)

remain = set(range(n))
ans = 0
while(len(remain) > 0):
    d = deque()
    d.append(remain.pop())
    num = 0
    xs = set()
    ys = set()
    while(len(d) > 0):
        tmp = d.pop()
        x,y = xy[tmp]
        num += 1
        xs.add(x)
        ys.add(y)
        while(len(x_list[x]) > 0):
            j = x_list[x].pop()
            if( j in remain):
                d.append(j)
                remain.remove(j)
        while(len(y_list[y]) > 0):
            j = y_list[y].pop()
            if(j in remain):
                d.append(j)
                remain.remove(j)

    ans += len(xs) * len(ys) - num

print(ans)