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
n = int(input())
xy=[list(map(int,input().split())) for i in range(n)]
from collections import defaultdict
x_d = defaultdict(list)
y_d = defaultdict(list)
for i in range(n):
    x,y = xy[i]
    x_d[x].append(i)
    y_d[y].append(i)

class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n
    def find(self, x):
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]
    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)
        if x == y:
            return
        if self.parents[x] > self.parents[y]:
            x, y = y, x
        self.parents[x] += self.parents[y]
        self.parents[y] = x
    def size(self, x):
        return -self.parents[self.find(x)]
    def same(self, x, y):
        return self.find(x) == self.find(y)
    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]
    def roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]
    def group_count(self):
        return len(self.roots())
uf = UnionFind(n)
for k in x_d.keys():
    if len(x_d[k])==1:
        continue
    last=x_d[k][0]
    for itm in x_d[k]:
        uf.union(last, itm)
        last=itm

for k in y_d.keys():
    if len(y_d[k])==1:
        continue
    last=y_d[k][0]
    for itm in y_d[k]:
        uf.union(last, itm)
        last=itm
ans=0
for root in uf.roots():
    mems = uf.members(root)
    xset=set()
    yset=set()
    for m in mems:
        x,y=xy[m]
        xset.add(x)
        yset.add(y)
    points = len(xset)*len(yset)
    points -= len(mems)
    ans+=points
print(ans)