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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

public class Main {
    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        int N = Integer.parseInt(sc.next());
        int[][] numArray = new int[10][10];
        for(int i=0;i<10;i++){
            for(int j=0;j<10;j++){
                numArray[i][j] = 0;
            }
        }
        for(int i=1;i<=N;i++){
            int topNum = Integer.parseInt(String.valueOf(i).substring(0,1));
            int bottomNum = i % 10;
            numArray[topNum][bottomNum]++;
        }
        int res = 0;
        for(int i=1;i<=N;i++){
            int topNum = Integer.parseInt(String.valueOf(i).substring(0,1));
            int bottomNum = i % 10;
            res += numArray[bottomNum][topNum];
        }
        System.out.println(res);
    }
}