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
def main():
    from collections import defaultdict
    import sys
    # functions used
    r = lambda: sys.stdin.readline().strip()
    R = lambda: list(map(int, r().split()))

    N = int(r())
    Dic = defaultdict(list)
    V = 10**5 + 5
    Arrived = [False] * (2 * V)
    for _ in range(N):
        x, y = R()
        y += V
        Dic[x].append(y)
        Dic[y].append(x)

    ans = 0
    task = []
    for k, v in Dic.items():
        if Arrived[k]:
            continue
        Arrived[k] = True
        if k < V:
            cntx = 1
            cnty = 0
        else:
            cntx = 0
            cnty = 1
        cntdot = 0
        task.extend(v)
        cntdot += len(v)
        while task:
            s = task.pop()
            if Arrived[s]:
                continue
            if s < V:
                cntx += 1
            else:
                cnty += 1
            Arrived[s] = True
            task.extend(Dic[s])
            cntdot += len(Dic[s])
        ans += cntx * cnty - cntdot // 2
    print(ans)

if __name__ == '__main__':
    main()
