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
N = int(input())

X = [[] for _ in range(10**5)]
Y = [[] for _ in range(10**5)]

P = [tuple(map(lambda x: int(x) - 1, input().split())) for _ in range(N)]

for i in range(N):
    x, y = P[i]
    X[x].append(y)
    Y[y].append(x)

x_used = [0 for _ in range(10**5)]
y_used = [0 for _ in range(10**5)]
res = 0

for i in range(10**5):
    if x_used[i]:
        continue
    stack = [(0, i)]
    x_cnt = 1
    x_used[i] = 1
    y_cnt = 0
    edge = 0
    while stack:
        axis, node = stack.pop()
        if axis == 0:
            for adj in X[node]:
                edge += 1
                if y_used[adj]:
                    continue
                y_cnt += 1
                y_used[adj] = 1
                stack.append((1, adj))
        else:
            for adj in Y[node]:
                if x_used[adj]:
                    continue
                x_cnt += 1
                x_used[adj] = 1
                stack.append((0, adj))
    res += x_cnt * y_cnt - edge

print(res)