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
def solve(points):
    N = len(points)
    X = defaultdict(list)
    Y = defaultdict(list)

    for i,(x,y) in enumerate(points):
        X[x].append(i)
        Y[y].append(i)
    
    res = 0
    visited = [False]*N
    for p0 in range(N):
        if visited[p0]:
            continue
        stack = [p0]
        visited[p0] = True
        xx = set()
        yy = set()
        while stack:
            p = stack.pop()
            x,y = points[p]
            xx.add(x)
            yy.add(y)

            for q in X[x]:
                if not visited[q]:
                    visited[q] = True
                    stack.append(q)
            for q in Y[y]:
                if not visited[q]:
                    visited[q] = True
                    stack.append(q)
        res += len(xx)*len(yy)
    return res-N




if __name__ == '__main__':
    N = int(input())
    points = [tuple(map(int,input().split())) for _ in range(N)]
    print(solve(points))