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: Wrong Answer)

#include <bits/stdc++.h>
#define ll long long

using namespace std;

int n;
ll sol=0;
ll cnt[10][10];
int a,b;

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);

	cin>>n;
    for(int j=1;j<n;j++)
    {
        int i=j;
        a=i%10;
        while(i!=0)
        {
            b=i;
            i/=10;
        }
        cnt[b][a]++;
    }
    for(int i=1;i<10;i++)
        for(int j=1;j<10;j++)
            sol+=cnt[i][j]*cnt[j][i];
    cout<<sol;

	return 0;
}