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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

n = int(input())

cnt = 0

for i in range(1, n+1):
    head = str(i)[0]
    tail = str(i)[-1]

    # 1桁
    if head == tail:
        cnt += 1

    if tail == "0":
        continue

    # 2桁
    if n <= 9:
        continue
    if int(tail + head) <= n:
        cnt += 1

    # 3桁
    if n <= 99:
        continue
    if int(tail + "0" + head) <= n:
        cnt += 10 if int(tail + "9" + head) < n else int(str(n)[1])

    # 4桁
    if n <= 999:
        continue
    if int(tail + "00" + head) <= n:
        cnt += 100 if int(tail + "99" + head) < n else int(str(n)[1:3])

    # 5桁
    if n <= 9999:
        continue
    if int(tail + "000" + head) <= n:
        cnt += 1000 if int(tail + "999" + head) < n else int(str(n)[1:4])

    # 6桁
    if n <= 99999:
        continue
    if int(tail + "0000" + head) <= n:
        cnt += 10000 if int(tail + "9999" + head) < n else int(str(n)[1:5])


print(cnt)