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
from bisect import bisect_left

N = int(input())
e5 = 10 ** 5

v = [[] for _ in range(2 * e5)]
visited = [True] * (2 * e5)

for _ in range(N) :
    x, y = map(int, input().split())
    x, y = x - 1, y - 1
    v[x].append(y + e5)
    v[y + e5].append(x)
    visited[x] = visited[y + e5] = False

ret = 0
for i in range(e5) :
    if not visited[i] :
        queue = [i]
        visited[i] = True
        nx, ny = 1, 0
        
        while queue :
            q = queue.pop()
            for j in v[q] :
                if not visited[j] :
                    visited[j] = True
                    queue.append(j)
                    
                    if j < e5 :
                        nx += 1
                    else :
                        ny += 1
                    
        ret += nx * ny
                    
print(ret - N)