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 deque
    import sys
    input = sys.stdin.readline

    M = 10 ** 5

    N = int(input())

    g = tuple(set() for _ in range(M * 2))
    for _ in range(N):
        x, y = (int(x) - 1 for x in input().split())  # 一律に移動
        g[x].add(y + M)
        g[y + M].add(x)

    visited = [0] * (M * 2)
    ans = 0
    for s in range(M * 2):
        if visited[s]: continue
        visited[s] = 1
        dq = deque()
        dq.append(s)

        sz = 0
        x = 0
        e = 0
        while dq:
            v = dq.popleft()
            sz += 1
            if v < M:
                x += 1
            for u in g[v]:
                e += 1
                if visited[u]: continue
                visited[u] = 1
                dq.append(u)
        ans += (sz - x) * x - (e // 2)
        # ans += (sz - x) * x - (sz - 1) # sz-1は木に限られるのでWA

    print(ans)


if __name__ == '__main__':
    main()
