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.chomp.to_i
a = Array.new(100,0)
b = 1
for i in 1..N do
    b = i >= 10**b ? b+1 : b
    min = i%10
    max = i/(10**(b-1))
    x = max * 10 + min
    a[x] += 1
end

answer = 0
for i in 1..9 do
    x = i*11
    answer += a[x]*a[x]
end
for i in 1..9 do
    for j in 1..9 do
        if i != j
            x = i*10+j
            y = i+j*10
            answer += a[x] * a[y]
        end
    end
end
puts answer