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

ans = 0

for i in 1..N do
  next if i.modulo(10) == 0
  if i.to_s[0] == i.to_s[-1] then
    n = i.to_s[0].to_i
  else
    n = (i.to_s[-1] + i.to_s[0]).to_i
  end
  while n <= N do
    if n.to_s.length == 1 then
      ans += 1
    elsif n > 100 && n.to_s[0] == N.to_s[0] && n.to_s.length == N.to_s.length then
      tmp = n
      while tmp <= N do
        ans += 1
        tmp += 10
      end
    else
     ans += 10 ** (n.to_s.length - 2)
     end
    n = (n.to_s[0] + "0" * (n.to_s.length - 1) + n.to_s[-1]).to_i
  end
end

puts ans