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)

#![allow(non_snake_case)]
#![allow(dead_code)]
#![allow(unused_macros)]
#![allow(unused_imports)]

use std::cmp::*;
use std::collections::*;

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};

    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
    ($iter:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    }
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };

    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };

    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };

    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

macro_rules! chmax {
    ($x:expr, $e:expr) => {
        if $x < $e {
            $x = $e;
        }
    };
}
macro_rules! chmin {
    ($x:expr, $e:expr) => {
        if $x > $e {
            $x = $e;
        }
    };
}

const INF: i64 = 1 << 60;

fn first_digit(mut n: usize) -> usize {
    while n >= 10 {
        n /= 10;
    }
    n
}
fn last_digit(n: usize) -> usize {
    n % 10
}

fn main() {
    input!{
        N: usize,
    }
    let mut m = HashMap::new();
    for i in 1..(N+1) {
        let e = m.entry((first_digit(i), last_digit(i))).or_insert(0);
        *e += 1;
    }
    let mut ans = 0;
    for i in 1..(N+1) {
        let f = first_digit(i);
        let l = last_digit(i);
        ans += match m.get(&(l, f)) {
            Some(&i) => i,
            None => 0,
        };
    }
    println!("{}", ans);
}