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
sys.setrecursionlimit(10**5)
m = 100005
visited = [False]*(m*2)
cnt = [0,0]
def dfs(i):
    visited[i] = True
    cnt[i//m] = cnt[i//m]+1
    for j in to[i]:
        if not visited[j]:
            dfs(j)
 
input = sys.stdin.readline
n = int(input())
to = [[] for _ in range(m*2)]
for _ in range(n):
    x,y = map(int,input().split())
    y = y+m
    to[x].append(y)
    to[y].append(x)
 
ans = 0
for i in range(m):
    if not visited[i]:
        cnt = [0,0]
        dfs(i)
        ans = ans + cnt[0]*cnt[1]
 
print(ans-n)