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
# coding=utf-8


get = lambda: int(raw_input().strip())
getl = lambda: raw_input().strip()
gets = lambda: map(int, raw_input().strip().split())
getss = lambda n: [gets() for _ in xrange(n)]


# N = 3
# A = [[1, 1], [5, 1], [5, 5]]

# N = 2
# A = [[10, 20], [20, 20]]

N = get()
A = getss(N)
UF = {}


def find(x):
    if x != UF[x]:
        UF[x] = find(UF[x])
    return UF[x]


def union(x, y):
    UF.setdefault(x, x)
    UF.setdefault(y, y)
    UF[find(x)] = find(y)


for i, j in A:
    union(i, -j)


count = {}
for x in UF:
    root = find(x)
    if root not in count:
        count[root] = [0, 0]
    if x > 0:
        count[root][0] += 1
    else:
        count[root][1] += 1

res = 0
for x in count:
    pos, neg = count[x]
    if pos > 1 and neg > 1:
        res += pos * neg
    else:
        res += max(pos, neg)
print res - N
