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 mod 1000000007
#define mod2 998244353 
#define PI 3.14159265
#define ll long long
#define ld long double
#define pi pair<ll,ll>
#define pb push_back
#define vi vector<ll>
#define sz size()
#define fi first
#define se second
#define lz length()
#define all(x) (x).begin(),(x).end()
#define scf(x) scanf("%lld",&x)
#define rep(i,n) for(ll i=0;i<n;i++)
const int INF = (int)1e9;
const ll IINF=1LL<<62;
const int maxn=10000005;
//int divr[maxn];
/**/
/*void divisor()
{for(int i=2;i<sqrt(maxn);i++)
     {
          if(!divr[i])
          {
               for(int j=i*i;j<maxn;j+=i)if(!divr[j])divr[j]=i;     
          }
     }
     for(int i=1;i<maxn;i++)if(!divr[i])divr[i]=i;
}*/
 
/*const int N=1000005;
ll n;
struct line {
    long long m, c;
    long long eval(long long x) { return m * x + c; }
    long double intersectX(line l) { return (long double) (c - l.c) / (l.m - m); }
};
vector<pair<pi,ll>> rect;
deque<line> dq;
*/
 
 
 
 


ll modexpo(ll a,ll b)
{
     ll res=1LL;
     while(b>0)
     {
          if(b&1)res=(res*a)%mod;
          a=(a*a)%mod;
          b>>=1;
     }
     return res;
}




















int main()
{
     int n;
     cin>>n;
     int c[10][10];
     memset(c,0,sizeof(c));
     for(int i=1;i<=n;i++){
          int t=i;
          int l=t%10;
          while(t>=10)
          {
               t/=10;
          }
          int f=t;
          c[f][l]++;

     }
     ll res=0LL;
     for(int i=1;i<=9;i++){
          for(int j=1;j<=9;j++){
               res+=c[i][j]*c[j][i];
          }
     }
     cout<<res;
}