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;
using lint = long long;
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#define rep(i, n) for(int i=0;i<(int)(n);++i)
typedef pair<int, int> P;

inline P func(int a) {
  int cnt = a % 10;
  while (a >= 10) {
    a/=10;
  }
  return {cnt, a};
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n; cin >> n;
  map<P, int> mp;
  for (int i = 1; i <= n; ++i) {
    P p = func(i);
    mp[p]++;
  }
  lint ans = 0;
  for (int i = 1; i <= n; ++i) {
    P p = func(i);
    P q = {p.second, p.first};
    ans += mp[q];
  }
  cout << ans << '\n';
}