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 LL long long int

int N, cnt[10][10];
LL answer;

int main() {
    cin >> N;

    for (int i = 1; i <= N; i++) {
        int number = i;

        int lastDigit = number%10;

        while (number >= 10) number /= 10;

        int firstDigit = number;

        cnt[firstDigit][lastDigit] += 1;
    }

    answer = 0;
    for (int f = 1; f <= 9; f++) {
        for (int s = 1; s <= 9; s++) {
            answer += ((LL) cnt[f][s] * cnt[s][f]);
        }
    }

    cout << answer << endl;
}