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)

/* Coached by Dukkha */
#include <stdio.h>

int first(int n) {
	return n < 10 ? n : first(n / 10);
}

int main() {
	static int kk[10][10];
	int n, x, y, ans;

	scanf("%d", &n);
	while (n) {
		x = first(n);
		y = n % 10;
		kk[x][y]++;
		n--;
	}
	ans = 0;
	for (x = 1; x <= 9; x++)
		for (y = 1; y <= 9; y++)
			ans += kk[x][y] * kk[y][x];
	printf("%d\n", ans);
	return 0;
}