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 int long long
#define double long double
#define SZ(x) ((int)(x).size())
#define rep(i,n) for(int i=0;i<(n);++i)
#define fore(i,a) for(auto &i:a)
#define min(...) min({__VA_ARGS__})
#define max(...) max({__VA_ARGS__})
#define Yes cout << "Yes" << endl;
#define No cout << "No" << endl;
#define ALL(x) (x).begin(),(x).end()
signed _main(); signed main() { cin.tie(0); ios::sync_with_stdio(false); _main(); }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> using v = vector<T>;
template<class T> using vv = vector<v<T>>;
const int MOD=1e9+7;
const long long INF = 1LL << 60;
using P=pair<int,int>;

signed _main(){
  int N,ans=0; cin >> N;
  vv<int> c(10,v<int>(10,0));
  for(int i=1;i<=N;i++){
    string s=to_string(i);
    c[s[0]-'0'][s[s.size()-1]-'0']++;
  }
  for(int i=0;i<=9;i++){
    for(int j=0;j<=9;j++){
      ans+=c[i][j]*c[j][i];
    }
  }
  cout << ans << endl;
  return 0;
}