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(200010)
n = int(input())
info = [list(map(int, input().split())) for i in range(n)]

v = 100000
graph = [[] for i in range(v*2)]
for i in range(n):
    x = info[i][0] - 1
    y = (info[i][1] - 1) + v
    graph[x].append(y)
    graph[y].append(x)

def dfs(pos):
    if visited[pos]:
        return
    visited[pos] = True
    cnt[pos//v] += 1
    for i in graph[pos]:
        dfs(i)
    return

visited = [False] * (v*2)
cnt = [0]*2
ans = 0
for i in range(v*2):
    cnt[0] = 0
    cnt[1] = 0
    dfs(i)
    ans += cnt[0]*cnt[1]
print(ans - n)
