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(object):
    def __init__(self, size):
        self.parent = [i for i in range(size)]
        self.rank = [0 for _ in range(size)]
        self.size = [1 for _ in range(size)]

    def find(self, x):
        if self.parent[x] == x:
            return x
        else:
            return self.find(self.parent[x])

    def unite(self, x, y):
        x = self.find(x)
        y = self.find(y)
        if x == y:
            return

        if self.rank[x] < self.rank[y]:
            self.parent[x] = y
            self.size[y]+=self.size[x]
        else:
            self.parent[y] = x
            self.size[x]+=self.size[y]
            if self.rank[x] == self.rank[y]:
                self.rank[x] += 1

    def same(self, x, y):
        return self.find(x) == self.find(y)
      
    def get_size(self, x):
        x = self.find(x)
        return self.size[x]



N = int(input())
U = UnionFind(N)
xyi = []
xy = []
for i in range(N):
    x, y = map(int, input().split())
    xyi.append((x, y, i))
    xy.append((x, y))
xyi.sort()
for j in range(N-1):
    x0, y0, i0 = xyi[j]
    x1, y1, i1 = xyi[j+1]
    if x0 == x1:
        U.unite(i0, i1)
xyi.sort(key = lambda x:x[1])
for j in range(N-1):
    x0, y0, i0 = xyi[j]
    x1, y1, i1 = xyi[j+1]
    if y0 == y1:
        U.unite(i0, i1)
equiv = [[set(), set()] for _ in range(N)]
for i in range(N):
    p = U.find(i)
    equiv[p][0].add(xy[i][0])
    equiv[p][1].add(xy[i][1])
ans = 0
for i in range(N):
    ans+=len(equiv[i][0])*len(equiv[i][1])
ans-=N
print(ans)





