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: Time Limit Exceeded)

import java.util.ArrayList;
import java.util.Scanner;

 
public class Main {
 
	public static void main(String[] args) {
		
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		ArrayList<Integer> AList = new ArrayList<>();
		ArrayList<Integer> BList = new ArrayList<>();
		int ans = 0;
		
		for(int i = 1; i <= N; i++) {
			AList.add(i);
			BList.add(i);
		}
		
		for(int i = 0; i < N; i++) {
			for(int j = 0; j < N; j++) {
				boolean firstCondition = true;
				boolean secondCondition = true;
				
				//A の末尾の桁が	 B  の先頭の桁に等しく、		 
				String A = AList.get(i).toString();
				Character Alast = A.charAt(A.length() - 1);
				Character Bfirst = BList.get(j).toString().charAt(0);
				if(!Alast.equals(Bfirst)) {
					firstCondition = false;
					continue;
				}
				//A の先頭の桁が	 B  の末尾の桁に等しい
				Character Afirst = AList.get(i).toString().charAt(0);
				String B = BList.get(j).toString();
				Character Blast = B.charAt(B.length() - 1);
				if(!Afirst.equals(Blast)) {
					secondCondition = false;
					continue;
				}
				
				if(firstCondition && secondCondition) {
					ans++;
				}
			}
		}
		
		System.out.println(ans);
		sc.close();
	}
}