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: Time Limit Exceeded)

N = gets.to_i
d = N.to_s.size # 桁数
array = ("1".."#{N}").to_a.reject { |n| n[-1] == "0" }
count = 0

# 整数"i...j"
(1..9).each do |i|
  (1..9).each do |j|
    if i == j
      if N < i
        next
      else
        m = array.size
        s = "^%s.*%s$" % [i, j]
        begin
          pair = (1 + (m - array.reject! { |n| n.match(s) }.size))**2
        rescue NoMethodError
          pair = 1
        end
        count += pair
      end
    else
      next if i > j
      
      m = array.size
      small, big = [i, j].sort
      s1 = "^%s.*%s$" % [small, big]
      s2 = "^%s.*%s$" % [big, small]
      begin NoMethodError
        num_s1 = m - array.reject! { |n| n.match(s1) }.size
        num_s2 = m - num_s1 - array.reject! { |n| n.match(s2) }.size
      rescue
        next
      end
      count += (num_s1 * num_s2) * 2
    end
  end
end

puts count