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 <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <climits>

using namespace std;

#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)

typedef long long LL;
typedef vector<int> VI;

int main() {
  int N;
  cin >> N;
  vector<vector<int>> c(10, vector<int>(10, 0));
  for (int i = 1; i <= N; i++) {
    string s = to_string(i);
    int l = s.at(0)-'0';
    int m = s.at(s.length()-1)-'0';
    c[l][m]++;
  }
  LL cnt = 0;
  for (int i = 0; i < 10; i++) {
    for (int j = 0; j < 10; j++) {
      cnt += c[i][j]*c[j][i];
    }
  }
  cout << cnt << endl;
  return 0;
}