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<cstdio>
#include<algorithm>
int i,j,k,n,x,a1,a2,r,a[10][10],p[]={1,10,100,1000,10000,100000};
void wk(int u)
{
	for(x=u;x;x/=10)a1=x%10;
	if(a1&&(a2=u%10))r+=a[a2][a1]+(a2*10+a1<=n)+(a1==a2);
}
int main()
{
	scanf("%d",&n);
	for(i=1;i<10;++i)for(j=1;j<10;++j)
		for(k=1;k<5;++k)
			for(x=0;x<p[k]&&(i*p[k]+x)*10+j<n;++x)++a[i][j];
	for(i=1;i<=n;++i)wk(i);
	printf("%d\n",r);
}