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: Accepted)

n = gets.to_i
n_digit = n.to_s.size
n_first = n.to_s[0].to_i
cnt = 0
(1..n).each do |i|
  i_digit = i.to_s.size
  i_first = i.to_s[0].to_i
  i_last = i.to_s[-1].to_i
  if i_first == i_last
    cnt += 1
  end
  next if i_last == 0
  k = 0
  loop do
    b = i_last * 10**(k+1) + i_first
    b_digit = b.to_s.size
    break if b > n
    if b_digit == n_digit && i_last == n_first
      cnt += (n - b) / 10 + 1
    else
      cnt += 10 ** k
    end
    k += 1
  end

end
puts cnt