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: Wrong Answer)

fun main(args: Array<String>) {
    val n = getInt()

    var r = 0L
    (1..n).forEach {
        val a = it.toString().first().toInt() - 48
        val b = it.toString().last().toInt() - 48
        if (b != 0) {
            r += num(b, a, n)
        }

        if (it == 1) {
            num(b, a, n)
        }
    }
    println(r)
}

fun num(a: Int, b: Int, n: Int): Int {
    var r = if (a == b) 1 else 0
    if (a * 10 + b <= n) r++
    else return r

    var digit = 1
    while (true) {
        val mini = min0(a, b, digit)
        val maxi = max0(a, b, digit)
        if (n >= maxi) {
            r += pow(10, digit)
        } else {
            if (n >= mini) {
                val rr = remain(n)
                if (rr == 0) r ++
                else r += rr
            }
            break
        }
        digit++
    }
    return r
}

val ten = arrayOf(10, 100, 1000, 10000, 100000, 1000000, 10000000)
val nine = arrayOf(0, 90, 990, 9990, 99990, 999990, 9999990)
fun max0(a: Int, b: Int, p: Int) = a * ten[p] + nine[p] + b
fun min0(a: Int, b: Int, p: Int) = a * ten[p] + b

fun pow(a: Int, p: Int): Int {
    var r = 1
    repeat(p) { r *= a }
    return r
}

fun remain(n: Int): Int {
    if (n <= 99) return 0
    return n.toString().drop(1).dropLast(1).toInt()
}

fun getInt() = readLine()!!.toInt()
fun getInts() = readLine()!!.split(" ").map { it.toInt() }
fun getLong() = readLine()!!.toLong()
fun getLongs() = readLine()!!.split(" ").map { it.toLong() }