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(arr:Array<String>) {
    val n = readLine()!!.toInt()
    val dic = (0..9).map { (0..9).map { 0 }.toMutableList() }
    for(i in (1..n)) {
        if(i % 10 == 0) {
            continue
        }
        val startNum = i.toString()[0].toString().toInt()
        val endNum = i % 10
        dic[startNum][endNum]++
    }
    var ans = 0L
    for(i in (1..9)) {
        for(j in (1..9)) {
            ans += dic[i][j].toLong() * dic[j][i].toLong()
        }
    }
    println(ans)
}