Problem:
Score : 400 points

Problem Statement

Given is a positive integer N.

Find the number of pairs (A, B) of positive integers not greater than N that satisfy the following condition:

When A and B are written in base ten without leading zeros, the last digit of A is equal to the first digit of B, and the first digit of A is equal to the last digit of B.

Constraints

1 \leq N \leq 2 \times 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N

Output

Print the answer.

Sample Input 1

25

Sample Output 1

17

The following 17 pairs satisfy the condition: (1,1), (1,11), (2,2), (2,22), (3,3), (4,4), (5,5), (6,6), (7,7), (8,8), (9,9), (11,1), (11,11), (12,21), (21,12), (22,2), and (22,22).

Sample Input 2

1

Sample Output 2

1

Sample Input 3

100

Sample Output 3

108

Sample Input 4

2020

Sample Output 4

40812

Sample Input 5

200000

Sample Output 5

400000008

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

n=gets.to_i
s=n.to_s.size
if s==1
  puts n
  exit
end

a=0
x=n/(10**(s-1))
y=n%10
z=(n%(10**(s-1)))/10

t=(10**(s-2)-1)/9
tt=10**(s-2)

1.upto(9) do |k|
  0.upto(9) do |l|
    if k==l
      c=1
      d=1
    else
      c=0
      d=0
    end
    c+=t
    d+=t
    if k<x
      c+=tt
    elsif k==x
      if l<=y
        c+=z+1
      else
        c+=z
      end
    end
    if l<x
      d+=tt
    elsif l==x
      if k<=y
        d+=z+1
      else
        d+=z
      end
    end
    a+=c*d
#    p [k,l,c,d,a]
  end
end
puts a