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)

macro_rules! scan_sub {
    ( $itr:ident, $t:ty ) => {
        let mut buf = String::new();
        let $itr = {
            while buf.trim().len() == 0 {
                buf = String::new();
                std::io::stdin().read_line(&mut buf).unwrap();
            }
            buf.split_whitespace().map(|n| n.parse::<$t>().unwrap())
        };
    };
}

macro_rules! scan {
    ( $x:ident; Vec<$t:ty> ) => {
        scan_sub!(itr, $t);
        let $x = itr.collect::<Vec<_>>();
    };
    ( $( $x:ident ),*; $t:ty ) => {
        scan_sub!(itr, $t);
        let mut itr = itr;
        $(
            let $x = itr.next().unwrap();
        )*
    };
}

fn main() {
    scan!(n; usize);
    let mut table = vec![vec![0; 10]; 10];

    if n < 10 {
        println!("{}", n);
        return;
    }

    for k in 1..(n+1) {
        let e = (k as f64).log10() as u32;
        let a = if e > 0 { k / 10_u32.pow(e) as usize} else { k };
        let b = k % 10;
        table[a][b] += 1;
    }

    let mut ans = 0;
    for i in 0..10 {
        for j in 0..10 {
            ans += table[i][j] * table[j][i];
        }
    }

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