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 <array>
using namespace std;

void f(int& start, int& end, int& k)
{
	if (k < 10) {
		start = k;
		end = k;
	}

	int temp = k;
	end = temp - temp / 10 * 10;
	temp /= 10;
	while (temp != 0) {
		start = temp - temp / 10 * 10;
		temp /= 10;
	}
}

int main()
{
	int n;
	cin >> n;

	array<array<int, 10>, 10> ary{0};

	int start, end;
	for (int i = 1; i <= n; ++i) {
		f(start, end, i);
		++ary[start][end];
	}

	long long ret = 0;
	for (int i = 1; i < 10; ++i) {
		for (int j = 1; j < 10; ++j) {
			ret += ary[i][j]* ary[j][i];
		}
	}

	cout << ret << endl;
	return 0;
}