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
from collections import defaultdict
import sys
input = sys.stdin.readline

class UnionFind:
    def __init__(self, N):
        self.parent = list(range(N))
        self.rank = [0] * N
        self.size = [1] * N

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

N = int(input())
a = [-1] * (10**5+1)
uf = UnionFind(10**5+1)
xs = [set() for _ in range(10**5+1)]
for _ in range(N):
    x, y = map(int, input().split())
    if a[x] == -1:
        a[x] = y
    else:
        uf.unite(a[x], y)
    xs[y].add(x)

root = defaultdict(set)
for y in range(10**5+1):
    root[uf.find(y)] |= xs[y]
ans = 0
for y in range(10**5+1):
    ans += len(root[uf.find(y)]) - len(xs[y])
print(ans)