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
sys.setrecursionlimit(10**7)

n = int(input())
xy = [[[] for _ in range(10**5+1)] for __ in range(2)]

for _ in range(n):
    x,y = map(int,input().split())
    xy[0][x].append(y)
    xy[1][y].append(x)

ans = 0
used = [[False for _ in range(10**5+1)] for __ in range(2)]

def dfs(i,flag):
    for v in xy[flag][i]:
        if not used[1-flag][v]:
            used[1-flag][v] = True
            dfs(v,1-flag)
            count[1-flag] += 1

for i in range(10**5+1):
    if xy[0][i] and not used[0][i]:
        used[0][i] = True
        count = [1,0]
        dfs(i,0)
        ans += count[0]*count[1]

ans -= n
print(ans)