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
data = Array.new(10){Array.new(10, 0)}

1.upto(9) do |i|
  if i <= n
    data[i][i] += 1
  end
  1.upto(9) do |j|
    next if i * 10 + j > n || j * 10 + i > n
    data[i][j] += 1
    cnt = 0
    keta = 1
    loop do
      num_str = sprintf("%0#{keta}d", cnt)
      num = (j.to_s + num_str + i.to_s).to_i
      break if num > n
      data[i][j] += 1
      if cnt < 10 ** keta -1 
        cnt += 1
      else
        keta += 1
        cnt = 0
      end
    end
  end
end

sum = 0
10.times do |i|
  i.upto(9) do |j|
    if i == j
      sum += data[i][j] ** 2
    else
      sum += 2 * data[i][j] * data[j][i]
    end
  end
end

p sum