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 Kotlin (1.0.0) (judge verdict: Accepted)

fun main(args: Array<String>) {
    val n = readLine()?.toInt() ?: return
    val pairs = 1.until(n + 1).map {
        val nn = it.toString().map { it.toInt() }
        val s = nn[0]
        val e = nn[nn.size - 1]
        Pair(s, e)
    }.groupBy { it }
    val memo = hashMapOf<Pair<Int, Int>, Int>()
    val count = pairs.keys.map {
        val p = Pair(Math.min(it.first, it.second), Math.max(it.first, it.second))
        memo[p] ?: run {
            val a = pairs[it]?.size ?: 0
            val b = pairs[Pair(it.second, it.first)]?.size ?: 0
            memo[p] = a * b
            a * b
        }
    }.sum()
    println(count)
}