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;
int main()
{
	int n;
	cin>>n;
	int sum = 0;
	for(int i=1;i<=9;i++)
	{
		for(int j=1;j<=9;j++)
		{
			int mid1=0;
			int mid2=0;
			for(int k=1;k<=n;k++)
			{
				int h=k,t=k%10;
				if(k>=10)
				{
					k/=10;
				}
				if (h == i && t == j)
					mid1++;
				if (h == j && t == i)
					mid2++;
			}
			sum += mid1 * mid2;
		}
	}
	cout<<sum<<endl;
	return 0;
}