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 (2.7.1) (judge verdict: Accepted)

n = gets.chomp.to_i

# 先頭がi, 末尾がjの自然数の個数
arr = 10.times.map{Array.new(10, 0)}

(1..n).each do |num|
  num_s = num.to_s
  arr[num_s[0].to_i][num_s[-1].to_i] += 1
end

count = 0
10.times do |i|
  10.times do |j|
    # A: arr[i][j], B: arr[j][i]とみなせば組み合わせは掛け算で表される
    count += arr[i][j] * arr[j][i]
  end
end

puts count