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)

N = gets.to_i
array = ("1".."#{N}").to_a.reject { |n| n[-1] == "0" }
cij = Array.new(9) { Array.new(9, 0) }
count = 0

array.each do |n|
    if n.to_i < 10
        cij[n.to_i - 1][n.to_i - 1] += 1
    else
        cij[n[0].to_i - 1][n[-1].to_i - 1] += 1
    end
end

9.times do |i|
    9.times do |j|
        count += cij[i][j] * cij[j][i]
    end
end

puts count