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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <math.h>

int digit(int n) {
	return (int)log10(n);
}

int main() {
	int N;
	long long int ans = 0;
	scanf("%d", &N);
	for (int i = 1; i <= 9; i++) {
		for (int j = 1; j <= 9; j++) {
			int A = 0;
			for (int a = 1; a <= N; a++) {
				if (a % 10 == j && a / (int)pow(10, digit(a)) == i){
					A++;
				}
			}
			int B = 0;
			for (int b = 1; b <= N; b++) {
				if (b % 10 == i && b / (int)pow(10, digit(b)) == j) {
					B++;
				}
			}
			ans = ans + (long long int)A * B;
		}
	}
	printf("%lld",ans);
	return 0;
}