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
sys.setrecursionlimit(10**6)
def input(): return sys.stdin.readline().strip()

def main():

    """
    <UnionFindを使わない別解>

    x方向とy方向に分けて頂点を辺だと思うのは定石？？？（確かに長方形がメインなら直積分解したくなる？）
    このとき問題文は「長さ３のパスがあればパスを１本追加する操作を繰り返して、各連結成分を完全二部グラフにせよ」

    完全二部グラフになるのは背理法からすぐわかる。なので解法は(¥Sum_{i:component}X_i Y_i) - N
    """

    N = int(input())
    V = 10 ** 5 + 1 # padding for y
    repn = [[] for _ in range(V * 2)]
    for _ in range(N):
        x, y = map(int, input().split())
        repn[x].append(V + y)
        repn[V + y].append(x)

    visited = [0] * (V * 2)
    count = [0, 0]

    # dfsでその都度連結成分を求める
    def dfs(v):
        if visited[v]:
            return
        visited[v] = 1
        count[v // V] += 1
        for u in repn[v]:
            dfs(u)

    ans = 0
    for i in range(V * 2):
        if visited[i]:
            continue
        count[:] = [0, 0]
        dfs(i)
        ans += count[0] * count[1]
    print(ans - N)


if __name__ == "__main__":
    main()
