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 Ruby (2.3.3) (judge verdict: Accepted)

def c(n, i, j)
	k = 1
	digits = 1
	count = 0
	while true
		k = 10 ** (digits - 1) * i
		while k / (10 ** (digits - 1)) == i and k <= n
			count += 1 if k % 10 == j
			k += 1
		end
		break if k > n

		digits += 1
	end
	count
end

def handstand(n)
	array = Array.new(9) {Array.new(9, -1)}
	count = 0
	(1..9).each do |i|
		(1..9).each do |j|
			if array[i - 1][j - 1] == -1
				array[i - 1][j - 1] = c(n, i, j)
			end
			if array[j - 1][i - 1] == -1
				array[j - 1][i - 1] = c(n, j, i)
			end
			count += array[i - 1][j - 1] * array[j - 1][i - 1]
		end
	end
	count
end

n = gets.chomp.to_i
puts handstand(n)