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>

int top(int x){
   while (x>=10) x/=10;
   return x;
}

int main(){
   int n,i,a,b,j;
   long ans;
 
   scanf("%d",&n);
   ans=0;
   for (i=n;i>0;i--) {
      a=top(i);
      b=i%10;
      if (b==0) continue;
      j=b*100000+a;
      if (j+99990<=n) ans+=10000;
      else if (j<=n) ans+=(n-j+10)/10;
      j=b*10000+a;
      if (j+9990<=n) ans+=1000;
      else if (j<=n) ans+=(n-j+10)/10;
      j=b*1000+a;
      if (j+990<=n) ans+=100;
      else if (j<=n) ans+=(n-j+10)/10;
      j=b*100+a;
      if (j+90<=n) ans+=10;
      else if (j<=n) ans+=(n-j)/10;
      j=b*10+a;
      if (j<=n) ans++;
      if (a==b) {
         j=a;
         if (j<=n) ans++;
      }
   }
   printf("%ld\n",ans);
   return 0;
}