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(100000)

N = int(input())
vertex = set()
conn = [[] for _ in range(2*10**5 + 1)]
check = [False]*(2*10**5 + 1)
for i in range(N):
    x, y = [int(x) for x in input().split()]
    vertex.add(x)
    vertex.add(y + 10**5)
    conn[x].append(y + 10**5)
    conn[y + 10**5].append(x)

count = [0] * 2
def dfs(z):
    if check[z] == True:
        return
    check[z] = True
    if z <= 10**5:
        count[0] += 1
    else:
        count[1] += 1
    for w in conn[z]:
        dfs(w)

ans = 0
for z in vertex:
    if check[z] == False:
        count = [0]*2
        dfs(z)
        X, Y = count[0], count[1]
        ans += X * Y
ans -= N

print(ans)