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::io::{self, BufRead};

fn main() {
  let stdin = io::stdin();
  let mut lines = stdin.lock().lines();
  let line = lines.next().unwrap().unwrap();
  let n: i32 = line.parse().unwrap();
  let mut memo = vec![vec![0; 9]; 9];
  for i in 1..(n+1) {
    let si = format!("{}", i);
    let z: usize = (0..(si.len()-1)).map(|_| 10).product();
    let most: usize = i as usize / z;
    let least: usize = i as usize % 10;
    if least == 0 {
      continue;
    }
    memo[most-1][least-1] += 1;
  }
  let mut count = 0;
  for i in 0..9 {
    for j in 0..9 {
      count += memo[i][j] * memo[j][i];
    }
  }
  println!("{}", count);
}