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
from collections import deque
from operator import itemgetter
input = sys.stdin.readline

def main():
    n = int(input())
    p = [list(map(int, input().split()))+[i] for i in range(n)]
    tree = [[] for _ in range(n)]
    
    p.sort()
    for i in range(n-1):
        if p[i][0] == p[i+1][0]:
            tree[p[i][2]].append(p[i+1][2])
            tree[p[i+1][2]].append(p[i][2])

    p.sort(key=itemgetter(1))
    for i in range(n-1):
        if p[i][1] == p[i+1][1]:
            tree[p[i][2]].append(p[i+1][2])
            tree[p[i+1][2]].append(p[i][2])

    p.sort(key=itemgetter(2))

    
    already = [False]*n
    ans = -n
    for i in range(n):
        if already[i]:
            continue
        already[i] = True
        not_yet = deque([i])
        key_x = {p[i][0]}
        key_y = {p[i][1]}
        while not_yet:
            key = not_yet.popleft()
            for v in tree[key]:
                if already[v]:
                    continue
                already[v] = True
                key_x.add(p[v][0])
                key_y.add(p[v][1])
                not_yet.append(v)
        ans += len(key_x)*len(key_y)
    
    print(ans)
        
    
    
if __name__ == "__main__":
    main()

