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
from collections import defaultdict
import sys


sys.setrecursionlimit(10 ** 6)


def main():
    N = int(input())
    V = 100005
    to = defaultdict(list)
    for _ in range(N):
        X, Y = map(int, input().split())
        Y += V
        to[X].append(Y)
        to[Y].append(X)
    visited = [0] * (2 * V)
    cnt = [0] * 2  # cnt = [cnt of X, cnt of Y]

    def dfs(v):
        if visited[v] == 1:
            return
        visited[v] = 1
        cnt[v // V] += 1
        for nv in to[v]:
            dfs(nv)

    ans = 0
    for v in range(2 * V):
        if visited[v] == 1:
            continue
        cnt = [0] * 2
        dfs(v)
        ans += cnt[0] * cnt[1]
    ans -= N

    print(ans)


if __name__ == '__main__':
    main()