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++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
bool is(long long i,long long j)
{
  int a1,a2,b1,b2;
  a1=i%10,b1=j%10;
  while(i)
  {
    a2=i%10;
    i /=10;
  }
   while(j)
  {
    b2=j%10;
    j /=10;
  }
  if(a1==b2)
  {
    if(a2==b1)
      return true;
    return false;
  }
  return false;
}
int main()
{
  long long n,count=0;
  cin>>n;
  for(long long i=1;i<=n;i++)
    for(long long j=1;j<=n;j++)
    {
         if(is(i,j))
         {
           count++;
         }
    }
  cout<<count;
  return 0;
}