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: Accepted)

n = int(input())

digits = 1
base = 10
for i in range(0, 20):
    if n < 10 ** i:
        digits = i
        break

smallest_digit = n % 10
largest_digit = n // (10 ** (digits - 1)) % 10
mid_digit_pattern = (n - largest_digit * (10 ** (digits - 1))) // 10 + 1

def count_pattern(a, b, dig):
    if dig == 1 and a != b:
        return 0
    elif dig == 1:
        return 1
    elif dig == digits:
        if a > largest_digit:
            return 0
        elif a < largest_digit:
            return 10 ** (dig - 2)
        elif b <= smallest_digit:
            return mid_digit_pattern
        elif b > smallest_digit:
            return mid_digit_pattern - 1
        else:
            assert False, "Can't reach here"
    else:
        return 10 ** (dig - 2)

result = 0
if digits == 1:
    result = n
else:
    counts = {}
    for i in range(1, 10):
        for j in range(1, 10):
            state = (i, j)
            counts[state] = sum([count_pattern(i, j, d) for d in range(1, digits + 1)])

    for state in counts.keys():
        result += counts[state] * counts[tuple(reversed(state))]

print(result)