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 Rust (1.15.1) (judge verdict: Time Limit Exceeded)

fn main() {
  let mut input = String::new();
  std::io::stdin().read_line(&mut input).ok();
  let n = input.trim().parse::<i32>().unwrap();
  let mut valid_num_cnt = 0;
  for i in 1..(n + 1) {
    for j in 1..(n + 1) {
      if get_last(i) == get_head(j) && get_head(i) == get_last(j) {
        valid_num_cnt += 1;
      }
    }
  }
  println!("{}",valid_num_cnt);
}

fn get_head(num:i32) -> i32 {
  if num <10 {
    num
  } else if num <100 {
    num / 10
  } else if num <1000 {
    num / 100
  } else if num <10000 {
    num / 1000
  } else if num <100000 {
    num / 10000
  } else {
    num / 100000
  }
}

fn get_last(num:i32) -> i32 {
  num % 10
}