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: Accepted)

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<queue>
#include<stack>
#include<math.h>
#include<map>
#include<string.h>
typedef long long int ll;
using namespace std;
int  d[100][100];
int mod=1e9+7;
int h(int x)
{
	return x%10;
}
int r(int x)
{   int s;
	while(x)
	{
		s=x%10;
		x/=10;
	}
   return s;
}
int main()
{
	int n,m,i,j;
	cin>>n;
	for(i=1;i<=n;i++)
	{
		d[h(i)][r(i)]++;
	}
	ll s=0;
	for(i=1;i<=9;i++)
	for(j=1;j<=9;j++)
	{
		s+=d[i][j]*d[j][i];
	}
	printf("%lld\n",s);
}