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 <string.h>
#include <stdlib.h>

int main(void)
{
    int n;
    int a, b;
    int c = 0;
    char sa[7], sb[7];
    int a2, a3, b2, b3;
    scanf("%d", &n);

    for (a = 1; a != n + 1; a++)
    {
        sprintf(sa, "%d", a);
        a2 = atoi(&sa[strlen(sa) - 1]);
        sa[1] = '\0';
        a3 = atoi(&sa[0]);
        for (b = 1; b != n + 1; b++) 
        {
            sprintf(sb, "%d", b);
            b3 = atoi(&sb[strlen(sb) - 1]);
            sb[1] = '\0';
            b2 = atoi(&sb[0]);
            if (a2 == b2 && a3 == b3)
            {
                c++;
            }
        }
    }
    printf("%d", c);
    return 0;
}