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)
v = 100005
visited = [False]*(2*v)
count = [0, 0]

def dfs(a):
    visited[a] = True
    count[a//v] += 1
    for i in vertex[a]:
        if not visited[i]:
            dfs(i)

input = sys.stdin.readline
N = int(input())
vertex = [[] for _ in range(v*2)]
for _ in range(N):
    x, y=map(int, input().split())
    y += v
    vertex[x].append(y)
    vertex[y].append(x)

ans = 0
for i in range(v):
    if not visited[i]:
        count = [0, 0]
        dfs(i)
        ans += count[0]*count[1]

print(ans-N)