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 DisjointSet:
    def __init__(self, size):
        self.rank = [0 for i in range(size)]
        self.p = [0 for i in range(size)]
        for i in range(size):
            self.makeSet(i)
    
    def makeSet(self, x):
        self.p[x] = x
        self.rank[x] = 0

    def same(self, x, y):
        return self.findSet(x) == self.findSet(y)

    def unite(self, x, y):
        self.link(self.findSet(x), self.findSet(y))

    def link(self, x, y):
        if self.rank[x] > self.rank[y]:
            self.p[y] = x
        else:
            self.p[x] = y
            if self.rank[x] == self.rank[y]:
                self.rank[y] += 1
    
    def findSet(self, x):
        if x != self.p[x]:
            self.p[x] = self.findSet(self.p[x])
        return self.p[x]
        

if __name__ == '__main__':
    maxxy = 10**5
    N = int(input())
    ds = DisjointSet(maxxy*2+1)
    left = [0 for i in range(maxxy*2+1)]
    right = [0 for i in range(maxxy*2+1)]
    for i in range(N):
        x, y = map(int, input().split())
        ds.unite(x, maxxy+y)
    for i in range(maxxy+1):
        left[ds.findSet(i)] += 1
    for i in range(maxxy+1, maxxy*2+1):
        right[ds.findSet(i)] += 1
    ans = 0
    for i in range(maxxy*2+1):
        ans += left[i] * right[i]
    ans -= N
    print(ans)
    