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 ** 7)
n = int(input())
edges = [[] for i in range(200000)]
for i in range(n):
    x, y = map(int, input().split())
    edges[x - 1].append(y + 99999)
    edges[y + 99999].append(x - 1)
ans = 0
visited = [0] * (200000)
cntx, cnty = 0, 0
def dfs(u):
    global cntx, cnty
    visited[u] = 1
    if u // 100000 >= 1:
        cnty += 1
    else:
        cntx += 1
    for i in edges[u]:
        if visited[i] == 0:
            dfs(i)
ans = -1 * n
for i in range(100000):
    if visited[i] == 0:
        cntx, cnty = 0, 0
        dfs(i)
        ans += cntx * cnty
print(ans)