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
input = sys.stdin.readline
sys.setrecursionlimit(1000000)
N = int(input())
K = 10**5+1
P = [i for i in range(K)]

def par(a):
    if P[a] == a:
        return a
    t = par(P[a])
    P[a] = t
    return t
def cb(a, b):
    pa = par(a)
    pb = par(b)
    if pa != pb:
        P[pa] = pb

X = [[] for _ in range(K)]
Y = [[] for _ in range(K)]
for _ in range(N):
    x, y = map(int, input().split())
    X[x-1].append(y-1)
    Y[y-1].append(x-1)
for i in range(K):
    for j in range(len(Y[i])-1):
        cb(Y[i][j], Y[i][j+1])

M = [[] for _ in range(K)]
for i in range(K):
    M[par(i)].append(i)
ans = -N
for i in range(K):
    S = []
    for j in M[i]:
        for k in X[j]:
            S.append(k)
    S = list(set(S))
    ans += len(S)*len(M[i])
print(ans)