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)

#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
using namespace std;
struct __INIT{__INIT(){cin.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(15);}} __init;
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define MOD 1000000007
#define INF (1<<30)
#define LINF (lint)(1LL<<56)
#define endl "\n"
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define reprev(i,n) for(int (i)=(n-1);(i)>=0;(i)--)
#define Flag(x) (1<<(x))
#define Flagcount(x) __builtin_popcount(x)
#define pint pair<int,int>
#define pdouble pair<double,double>
#define plint pair<lint,lint>
typedef long long lint;
int dx[8]={1,1,0,-1,-1,-1,0,1};
int dy[8]={0,1,1,1,0,-1,-1,-1};

lint calc(lint x){
    lint keta=(lint)(1+floor(log10(x)));
    lint top=x/(lint)(pow(10,keta-1));
    lint res=0;
    lint ketac=2;
    lint botom=x%10;
    if(top==botom){
        res+=3;
        while(ketac<keta) res+=(lint)pow(10,ketac-2)*2,ketac++;
        lint rem=(x-top*(lint)pow(10,keta-1)-botom)/5;
        if(keta>=3) res+=rem;
    }
    if(top>botom) while(ketac<=keta) res+=(lint)pow(10,ketac-2)*2,ketac++;
    if(top<botom) while(ketac<keta) res+=(lint)pow(10,ketac-2)*2,ketac++;
    return res;
}

int main(void){
    lint N;
    cin >> N;
    lint dp[220000]={};
    for(int i=1;i<=9;i++) dp[i]=i;
    for(int i=10;i<=N;i++){
        if(i%10==0) dp[i]=dp[i-1];
        else dp[i]=dp[i-1]+calc(i);
    }
    cout << dp[N] << endl;
}