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 f_must_be_rectangular(N, Positions):
    # 参考: https://atcoder.jp/contests/abc131/submissions/6086451
    # x軸, y軸をまとめて1つのリストで扱う
    import sys
    sys.setrecursionlimit(10**6)

    vertex_maxsize = 10**5 + 5
    target = [[] for _ in range(2 * vertex_maxsize)]
    is_visited = [False] * (2 * vertex_maxsize)
    for x, y in Positions:
        y += vertex_maxsize  # 1つのリストで扱う都合上、ずらす
        target[x].append(y)
        target[y].append(x)

    def dfs(v):
        if is_visited[v]:
            return
        is_visited[v] = True
        count[v // vertex_maxsize] += 1  # vがx軸とy軸のどっちを指しているか
        for u in target[v]:
            dfs(u)  # 頂点vに接続されている頂点へ向かう

    ans = 0
    for i in range(2 * vertex_maxsize):
        if is_visited[i]:
            continue
        count = [0] * 2
        dfs(i)  # 頂点iにつながっている頂点を調べる
        # x座標とy座標のそれぞれについて、ペアにすると矩形が完成する
        # 座標がいくつあるか。その積が解となる
        ans += count[0] * count[1]
    return ans - N  # 最初に与えられた座標の分、引く

N = int(input())
Positions = [[int(i) for i in input().split()] for j in range(N)]
print(f_must_be_rectangular(N, Positions))