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
class UnionFind:
    def __init__(self, n):
        self.n = n
        self.parent = [None]*n

    def root(self, x):
        if self.parent[x] == None:
            return x
        else:
            return self.root(self.parent[x])

    def union(self, x, y):
        x0 = self.root(x)
        y0 = self.root(y)
        if x0 == y0:
            return
        if x0 > y0:
            x0, y0 = y0, x0
        self.parent[y0] = x0


n = int(input())
m = 10**5
uf = UnionFind(2*m)
X, Y = [], []
for i in range(n):
    x, y = map(int, input().split())
    X.append(x-1)
    Y.append(y-1+m)
    uf.union(x-1, y-1+m)
R = {}
for x, y in zip(X, Y):
    r = uf.root(x)
    try:
        R[r][0].add(x)
        R[r][1].add(y)
    except:
        R[r] = [{x}, {y}]
ans = 0
for r in R.keys():
    ans += len(R[r][0])*len(R[r][1])
print(ans-n)
