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 (2.7.1) (judge verdict: Accepted)

# frozen_string_literal: true

n = gets.to_i
nn = n
digits = []
while nn != 0
  digits << nn % 10
  nn /= 10
end
ketac = digits.size
msd = digits.last
table = (1..9).map { |i| Array.new(9, 0) }
(1..9).each do |signi|
  (1..9).each do |lesig|
    tmp = (2...ketac).sum do |d|
      10**(d - 2)
    end + (signi == lesig && lesig <= n ? 1 : 0)
    case signi <=> msd
    when -1
      tmp += 10**(ketac - 2)
    when 0
      unless ketac < 2
        full = digits[1...-1].reverse!.inject(0) { |prod, d| prod * 10 + d }
        full = 0 if digits[1..-2].empty?
        full += 1 if lesig <= digits[0]
        tmp += full
      end
    end
    table[signi - 1][lesig - 1] = tmp
  end
end
res = (0..8).sum do |i|
  (0..8).sum do |j|
    table[i][j] * table[j][i]
  end
end
p res