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)

fn calc_msd(x: usize) -> usize {
    let mut msd = x;
    while msd >= 10 {
        msd /= 10;
    }
    msd
}

fn calc_comb(n: usize) -> Vec<Vec<u64>> {
    let mut comb = vec![vec![0u64; 10]; 10];
    for x in 1..(n+1) {
        let msd = calc_msd(x);
        let lsd = x%10;
        comb[msd][lsd] += 1;
    }
    comb
}

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_line(&mut buf).unwrap();
    let mut iter = buf.split_whitespace();
    let n: usize = iter.next().unwrap().parse().unwrap();

    let comb = calc_comb(n);

    let mut ans = 0u64;
    for i in 1..10 {
        for j in 1..10 {
            ans += comb[i][j] * comb[j][i];
        }
    }

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