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)

def first_and_last_num_by(n)
  first = n % 10
  last = n
  while n > 0
    last = n
    n /= 10
  end
  [first, last]
end

n = gets.to_i

count = Hash.new(0)
(1..n).each do |e|
  first, last = first_and_last_num_by(e)
  key = [first, last].join(',')
  count[key] += 1
end

ans = 0
(1..n).each do |e|
  first, last = first_and_last_num_by(e)
  key = [last, first].join(',')
  ans += count[key]
end

puts ans