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
#!/usr/bin/env python3
import sys
from collections import defaultdict
from collections import deque


def solve(N: int, X: "List[int]", Y: "List[int]"):
    adj = defaultdict(list)
    done = set()
    for x, y in zip(X, Y):
        xnode = (x, 0)
        ynode = (y, 1)
        adj[xnode].append(ynode)
        adj[ynode].append(xnode)

    points_to_add = 0    
    for x, y in zip(X, Y):
        xnode = (x, 0)
        ynode = (y, 1)
        if xnode in done or ynode in done:
            continue
        x_count = 1
        y_count = 0
        done.add(xnode)
        Q = deque()
        Q.append(xnode)
        while len(Q) > 0:
            node = Q.pop()
            for node_next in adj[node]:
                if node_next in done:
                    continue
                if node_next[1] == 0:
                    x_count += 1
                else:
                    y_count += 1
                Q.append(node_next)
                done.add(node_next)
        points_to_add += x_count * y_count
    print(points_to_add - N)
        
    return


# Generated by 1.1.4 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
def main():
    def iterate_tokens():
        for line in sys.stdin:
            for word in line.split():
                yield word
    tokens = iterate_tokens()
    N = int(next(tokens))  # type: int
    x = [int()] * (N)  # type: "List[int]" 
    y = [int()] * (N)  # type: "List[int]" 
    for i in range(N):
        x[i] = int(next(tokens))
        y[i] = int(next(tokens))
    solve(N, x, y)

if __name__ == '__main__':
    main()
