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::{stdin, Read};

macro_rules! parse {
    ($iter:ident) => {
        $iter.next().unwrap().parse().unwrap()
    };
    ($iter:ident, 2) => {
        (parse!(iter), parse!(iter))
    };
    ($iter:ident, 3) => {
        (parse!(iter), parse!(iter))
    };
    ($iter:ident, 4) => {
        (parse!(iter), parse!(iter))
    };
}

fn main() {
    let mut s = String::new();
    stdin().read_to_string(&mut s).unwrap();
    let mut iter = s.split_whitespace();
    let n: isize = parse!(iter);
    let mut se = vec![vec![0; 10]; 10];
    for mut i in 1..(n + 1) {
        let s = i % 10;
        while i >= 10 {
            i /= 10;
        }
        se[i as usize][s as usize] += 1;
    }
    let mut res = 0u64;
    for i in 0..10 {
        for j in 0..10 {
            res += se[i][j] * se[j][i]
        }
    }
    println!("{}", res);
}