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
table = Array.new(10){Array.new(10){0}}

(1..9).each do |f|
  (1..9).each do |e|
    if f == e
      n = f
      if n <= N
        table[f][e] += 1
      end
    end

    flag = true
    loop_count = 0
    while flag
      n = "#{f}#{"9" * loop_count}#{e}".to_i
      if n <= N
        table[f][e] += 10 ** loop_count
      else
        (("0" * loop_count)..("9" * loop_count)).each do |i|
          n = "#{f}#{i}#{e}".to_i
          if n <= N
            table[f][e] += 1
          else
            break
          end
        end
        flag = false
      end
      loop_count += 1
    end
  end
end

result = 0
(1..N).each do |n|
  f = n.to_s[0].to_i
  e = n.to_s[-1].to_i

  result += table[e][f]
end
puts result