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
read = sys.stdin.buffer.read
readlines = sys.stdin.buffer.readlines
input = sys.stdin.buffer.readline
sys.setrecursionlimit(10**7)


def main():
    OFS = 100005
    N = int(input())
    G = [[] for _ in [0]*OFS*2]
    for line in readlines():
        x, y = map(int, line.split())
        y += OFS
        G[x].append(y)
        G[y].append(x)

    def dfs(v):
        if v < OFS:
            counter[0] += 1
        else:
            counter[1] += 1

        for u in G[v]:
            if visited[u]:
                continue
            visited[u] = 1
            dfs(u)

    visited = [0]*OFS*2
    ans = 0
    for i in range(OFS):
        if visited[i]:
            continue
        counter = [0]*2
        visited[i] = 1
        dfs(i)
        ans += counter[0] * counter[1]

    print(ans - N)


if __name__ == '__main__':
    main()