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;

int main() {
	int n; cin >> n;
	vector<vector<int>> v(9, vector<int>(9, 0));
	for (int i = 1; i <= n; i++) {
		int j = i, x, y;
		x = j % 10;
		int d = 10;
		while (j / d > 0) d *= 10;
		d /= 10;
		y = j / d;
		if (x > 0) v.at(x - 1).at(y - 1)++;
	}
	int ans = 0;
	for (int i = 0; i < 9; i++) {
		for (int j = 0; j < 9; j++) {
			ans += v.at(i).at(j) * v.at(j).at(i);
		}
	}
	cout << ans << endl;
}