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: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
	int n;
	scanf("%d", &n);
	int cnt = 0, lenA;
	char a[7], b[7];
	for (int x = 1; x <= n; x++)
	{
		sprintf(a, "%d", x);
		lenA = (int)strlen(a) - 1;
		for (int y = x; y <= n; y++)
		{
			sprintf(b, "%d", y);
			if (a[0] == b[strlen(b) - 1] && a[lenA] == b[0])
			{
				if (x == y) cnt++;
				else cnt += 2;
			}
		}
	}
	printf("%d", cnt);
}