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;
	 
#define fast_io ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
#define cout_precision cout.setf(ios::fixed); cout.precision(32);
#define clr(a) memset(a,0,sizeof(a))
#define umap unordered_map
#define uset unordered_set
#define fr first
#define sc second
#define pb push_back
#define M int(1e9+7)
#define endl '\n'
	
using ll = int64_t; using vll = vector<ll>; using vvll = vector<vll>;
using pll = pair<ll, ll>; using vpll = vector<pll>; using vvpll = vector<vpll>;
	
	
int main(){
	fast_io; cout_precision;
	int n;	cin >> n;
	int dp[10][10];	clr(dp);
	for(int i=1; i<=n; i++){
		int k = (i/pow(10 , floor(log10(i))));
		dp[i%10][k]++;
	}
	int ans = 0;
	for(int i=1; i<10; i++){
		for(int j=1; j<10; j++)	ans+=dp[i][j]*dp[j][i];
	}
	cout << ans <<endl;
	return 0;
}