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: Accepted)

use std::collections::HashMap;
fn read<T: std::str::FromStr>() -> T {
  let mut s = String::new();
  std::io::stdin().read_line(&mut s).ok();
  s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
  read::<String>().split_whitespace()
    .map(|e| e.parse().ok().unwrap()).collect()
}

fn read_vec2<T: std::str::FromStr>(n: u32) -> Vec<Vec<T>> {
  (0..n).map(|_| read_vec()).collect()
}

fn main() {
  let n: u32 = read();
  let mut hash = HashMap::new();

  for i in 1..n+1 {
    let pair = pair(i);
    let count = hash.entry(pair).or_insert(0);
    *count += 1;
  }

  let mut ans: u64 = 0;

  for i in 1..n+1 {
    let pair = pair(i);
    let pair_pair = (pair.1, pair.0);
    match hash.get(&pair_pair) {
      Some(count) => ans += *count,
      None => ans += 0
    }
  }

  println!("{}", ans);
}

fn pair(i: u32) -> (u32, u32) {
  let last = i%10;
  let start: u32 = i.to_string().chars().nth(0).unwrap().to_digit(10).unwrap();
  return (start, last);
}