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-*--

def dsMakeSet(ds, x):
    ds[x] = x


def dsFind(ds, x):
    if ds[x] == x:
        return x

    x0 = x
    while ds[x] != x:
        x = ds[x]

    ds[x0] = x
    return x


def dsUnion(ds, x, y):
    xRoot = dsFind(ds, x)
    yRoot = dsFind(ds, y)
    
    if xRoot != yRoot:
        ds[yRoot] = xRoot
        return True

    return False



N = int(input())
Q = {}

for _ in range(N):
    x, y = map(int, input().split())

    X = ('x', x)
    Y = ('y', y)

    if not (X in Q):
        dsMakeSet(Q, X)

    if not (Y in Q):
        dsMakeSet(Q, ('y', y))
    
    dsUnion(Q, X, Y)

R = {}

for q in Q:
    root = dsFind(Q, q)
    
    if not(root in R):
        R[root] = [0, 0]

    r = R[root]
    if q[0] == 'x':
        r[0] += 1
    else:
        r[1] += 1

s = 0
for r in R.values():
    s += r[0]*r[1]

print(s - N)

    

