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


def dfs(v):
    cnt = [0, 0]
    if v < OFFSET:
        cnt[0] += 1
    else:
        cnt[1] += 1
    visited[v] = True
    stack = [v]
    while stack:
        v = stack.pop()
        for nxt_v in graph[v]:
            if visited[nxt_v]:
                continue
            if nxt_v < OFFSET:
                cnt[0] += 1
            else:
                cnt[1] += 1
            visited[nxt_v] = True
            stack.append(nxt_v)
    return cnt[0] * cnt[1]

n = int(input())
info = [list(map(int, input().split())) for i in range(n)]
OFFSET = 10 ** 5

graph = [[] for i in range(2 * OFFSET)]
for x, y in info:
    x -= 1
    y -= 1
    y += OFFSET
    graph[x].append(y)
    graph[y].append(x)

ans = 0
visited = [False] * (2 * OFFSET)
for v in range(2 * OFFSET):
    if visited[v]:
        continue
    else:
        ans += dfs(v)
print(ans - n)