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 Python (3.4.3) (judge verdict: Accepted)

N = int(input())

def count(a):
  if N < 10:
    if N < a:
      return 0
    else:
      return 1
  else:
    cnt_a = 1
    keta = 2
    while True:
      if N > int(str(a)+str(9)*(keta-2)+str(a)):
        cnt_a += 10 ** (keta-2)
        keta += 1
      elif N >= int(str(a)+str(0)*(keta-2)+str(a)):
        cnt_a += (N - a*(10**(keta-1)) - a) // 10 + 1
        break
      else:
        break
  return cnt_a
        
  
def count_(a, b):
  if N < 10:
    return 0, 0
  else:
    cnt_ = 0
    cnt__  = 0
    keta = 2
    while True:
      if N > int(str(b)+str(9)*(keta-2)+str(a)):
        cnt_ += 10 ** (keta-2)
        cnt__ += 10 ** (keta-2)
        keta += 1
      elif N >= int(str(b)+str(0)*(keta-2)+str(a)):
        cnt_ += (N - b*(10**(keta-1)) - a) // 10 + 1
        cnt__ += 10 ** (keta-2)
        break
      elif N > int(str(a)+str(9)*(keta-2)+str(b)):
        cnt_ += 10 ** (keta-2)
        break
      elif N >= int(str(a)+str(0)*(keta-2)+str(b)):
        cnt_ += (N - a*(10**(keta-1)) - b) // 10 + 1
        break
      else:
        break
  return cnt_, cnt__

ans = 0
for i in range(1, 10):
  for j in range(i, 10):
    # start, end  =  i, j
    if i == j:
      # １桁が可能
      #print(i, count(i))
      ans += count(i)**2
    else:
      pa, pa_ = count_(i,j)
      #print(i, j, pa, pa_)
      ans += pa * pa_ * 2
      
print(ans)