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 OCaml (4.02.3) (judge verdict: Accepted)

let rec fix f x = f (fix f) x
let () =
  Scanf.scanf "%d" @@ fun n ->
  let ad = Array.init 10 (fun _ -> Array.make 10 0) in
  for i = 1 to n do
    let d = fix (fun f x -> if x < 10 then x else f (x/10)) i in
    ad.(i mod 10).(d) <- ad.(i mod 10).(d) + 1;
  done;

  fix (fun f ans i j ->
    if i > 9 then ans
    else if j > 9 then f ans (i+1) 1
    else f (ans + ad.(i).(j) * ad.(j).(i)) i (j+1)) 0 1 1 
  |> Printf.printf "%d\n"