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<bits/stdc++.h>
using namespace std;
int n;
int find_p(int x) {
	int tmp=x;
	while(tmp>=10) tmp/=10;
	return tmp;
}
int ans=0,c[20][20];
int main()
{
	memset(c,0,sizeof(c));
	scanf("%d",&n);
	for(int i=1;i<=n;++i) {
		int j=find_p(i),k=i%10;
		c[j][k]++;
	}
	for(int i=0;i<=9;++i)
		for(int j=0;j<=9;++j)
			ans+=c[i][j]*c[j][i];
	printf("%d",ans);
	return 0;
}