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 REP(i, n) for (int i = 0; i < (int)(n); i++)
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
typedef long long ll;
typedef pair<int, int> P;
const int INF = 100100100;
const int MOD = (int)1e9 + 7;
const double EPS = 1e-9;
int main() {
  ll N;
  cin >> N;
  vector<vector<int>> A(10, vector<int>(10, 0));
  ll ans = 0;
  for (int i = 1; i <= N; i++) {
    if (i % 10 == 0) continue;
    string s = to_string(i);
    int a = s[0] - '0';
    if (a == 0) continue;
    A[a][i % 10]++;
  }
  for (int i = 1; i <= N; i++) {
    if (i % 10 == 0) continue;
    string s = to_string(i);
    int a = s[0] - '0';
    if (a == 0) continue;
    ans += A[i % 10][a];
  }
  cout << ans << endl;
}