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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.io.IOException;
import java.io.InputStream;
import java.util.Map;
import java.util.NoSuchElementException;

class FastScanner {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;

    private boolean hasNextByte() {
        if (ptr < buflen) {
            return true;
        } else {
            ptr = 0;
            try {
                buflen = in.read(buffer);
            } catch (IOException e) {
                e.printStackTrace();
            }
            if (buflen <= 0) {
                return false;
            }
        }
        return true;
    }

    private int readByte() {
        if (hasNextByte())
            return buffer[ptr++];
        else
            return -1;
    }

    private static boolean isPrintableChar(int c) {
        return 33 <= c && c <= 126;
    }

    public boolean hasNext() {
        while (hasNextByte() && !isPrintableChar(buffer[ptr]))
            ptr++;
        return hasNextByte();
    }

    public String next() {
        if (!hasNext())
            throw new NoSuchElementException();
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while (isPrintableChar(b)) {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }

    public long nextLong() {
        if (!hasNext())
            throw new NoSuchElementException();
        long n = 0;
        boolean minus = false;
        int b = readByte();
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        if (b < '0' || '9' < b) {
            throw new NumberFormatException();
        }
        while (true) {
            if ('0' <= b && b <= '9') {
                n *= 10;
                n += b - '0';
            } else if (b == -1 || !isPrintableChar(b)) {
                return minus ? -n : n;
            } else {
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }

    public int nextInt() {
        long nl = nextLong();
        if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE)
            throw new NumberFormatException();
        return (int) nl;
    }

    public double nextDouble() {
        return Double.parseDouble(next());
    }
}

public class Main {
    static int[] w = new int[100010];
    static int[] v = new int[100010];
    static int[][] dp = new int[110][100010];
    static int k;

    static int chmin(int a, int b) {
        if (a > b) {
            a = b;
            return a;
        }
        return a;
    }

    static int chmax(int a, int b) {
        if (a < b) {
            a = b;
            return a;
        }
        return a;
    }

    static int calcNumOfCombination(int n, int r) {
        int num = 1;
        for (int i = 1; i <= r; i++) {
            num = num * (n - i + 1) / i;
        }
        return num;
    }

    static long lcm(long a, long b) {
        long temp;
        long c = a;
        c *= b;
        while ((temp = a % b) != 0) {
            a = b;
            b = temp;
        }
        return (long) (c / b);
    }

    static long f(long n, Map<Long, Long> map) {
        if (n < 2)
            return 1;
        else if (map.containsKey(n)) {
            return map.get(n);
        } else {
            map.put(n, n * f(n - 2, map));
            return map.get(n);
        }

    }

    public long factorialUsingRecursion(int n) {
        if (n <= 2) {
            return n;
        }
        return n * factorialUsingRecursion(n - 1);
    }

    public static void main(String[] args) {
        FastScanner fs = new FastScanner();
        int n = fs.nextInt();
        int ans = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                char[] a = String.valueOf(i).toCharArray();
                char[] b = String.valueOf(j).toCharArray();
                if (a[0] == b[b.length - 1] && a[a.length - 1] == b[0])
                    ans++;
            }
        }

        System.out.println(ans);
    }

}