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)

def main():
    input = sys.stdin.readline
    V = 10**5  # max size of x
    N = int(input())
    bgraph = [[] for _ in range(2*V)]
    for _ in range(N):
        x, y = map(int, input().split())
        x -= 1
        y += V - 1
        bgraph[x].append(y)
        bgraph[y].append(x)

    def dfs(i):
        if visited[i]:
            return cnt
        visited[i] = True
        cnt[i//V] += 1
        for j in bgraph[i]:
            dfs(j)
        return cnt

    visited = [False] * (2*V)
    ans = 0
    for i in range(V):
        if visited[i]:
            continue
        cnt = [0, 0]
        cnt = dfs(i)
        ans += cnt[0] * cnt[1]

    ans -= N
    return ans


if __name__ == '__main__':
    print(main())
