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**6)

MAXVAL = 10**5
N = int(input())
edge = [[] for _ in range(MAXVAL*2)]
visited = [0] * (MAXVAL*2)
ans = 0
xnum = 0; ynum = 0; nodes = 0

def dfs(n):
    global xnum, ynum, nodes
    if visited[n] == 1:
        return True
    if n < MAXVAL:
        xnum += 1
    else:
        ynum += 1
    visited[n] = 1
    for item in edge[n]:
        nodes += 1
        dfs(item)
    return True

for i in range(N):
    x, y = [int(item) for item in input().split()]
    x -= 1; y -= 1
    edge[x].append(y+MAXVAL)
    edge[y+MAXVAL].append(x)

for i in range(MAXVAL*2):
    if visited[i] == 1:
        continue
    xnum = 0; ynum = 0; nodes = 0
    dfs(i)
    if nodes >= 3:
        ans += xnum * ynum - nodes // 2
print(ans)