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 dfs(x, al):
    global cntx, cnty
    if (al, x) in used:
        return
    used.add((al, x))
    if al == 0:
        cntx += 1
    else:
        cnty += 1
    for xy in to[al][x]:
        dfs(xy, 1 - al)
    return

n = int(input())
to = [{}, {}]
for _ in range(n):
    x, y = map(int, input().split())
    to[0].setdefault(x, [])
    to[1].setdefault(y, [])
    to[0][x] += [y]
    to[1][y] += [x]
used = set()
ans = 0
for x in to[0].keys():
    if (0, x) in used: continue
    cntx = 0
    cnty = 0
    dfs(x, 0)
    ans += cntx * cnty
print(ans - n)
