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: Accepted)

import static java.util.Comparator.comparing;

import java.io.InputStream;
import java.io.PrintWriter;
import java.math.BigDecimal;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Objects;
import java.util.PriorityQueue;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        PrintWriter out = new PrintWriter(System.out);
        Solver solver = new Solver(System.in, out);
        solver.solve();
        out.close();
    }
}
class Solver {
	Scanner sc;
	PrintWriter out;
    public Solver(InputStream in, PrintWriter out) {
    	sc = new Scanner(in);
    	this.out = out;
    }
    // ==========================================================
    Map<PPKEY, Integer> map = new HashMap<>();	// 先頭と末端の組の数を設定する
    // 先頭と末端の数字の組を作る
    PPKEY f(int N) {
    	int a = N % 10;
    	int b = N;
    	while(b >= 10) {
			b /= 10;
		}
//    	out.println("N = " + N + "  a = " + a + "  b = " + b);
    	return new PPKEY(a, b);
    }
    public void solve() {
    	int N = ni();
    	PPKEY wk;
    	for (int i = 1; i <= N; i++) {
			wk = f(i);
			map.merge(wk, 1, Integer::sum);
		}
    	long ans = 0;
    	Integer c;
    	for (int i = 1; i <= N; i++) {
			wk = f(i);
			wk = new PPKEY(wk.val, wk.key);
			c = map.get(wk);
			ans += (c == null ? 0 : c);
    	}
    	out.println(ans);
    }
    final class PPKEY{
    	private final int key, val;
    	public PPKEY(int key, int val) {
    		this.key = key;
    		this.val = val;
    	}
		public int getKey() {
			return key;
		}
		public int getVal() {
			return val;
		}
	    @Override
	    public boolean equals(Object obj) {
	        if (obj instanceof PPKEY) {
	        	PPKEY dest = (PPKEY) obj;
	            return this.key == dest.key && this.val == dest.val;
	        } else {
	            return false;
	        }
	    }
	    @Override
	    public int hashCode() {
	        return Objects.hash(key, val);
	    }
    }
    // ------------------------------------------
    // 入力
    // ------------------------------------------
    public int ni() {
    	return sc.nextInt();
    }
    public long nl() {
    	return sc.nextLong();
    }
    public String ns() {
    	return sc.next();
    }
    public char[] nc() {
    	return sc.next().toCharArray();
    }
    public int[] ndi(int N) {
    	int[] ans = new int[N];
    	for (int i = 0; i < N; i++) {
			ans[i] = ni();
		}
    	return ans;
    }
    public long[] ndl(int N) {
    	long[] ans = new long[N];
    	for (int i = 0; i < N; i++) {
			ans[i] = nl();
		}
    	return ans;
    }
    public String[] nds(int N) {
    	String[] ans = new String[N];
    	for (int i = 0; i < N; i++) {
			ans[i] = ns();
		}
    	return ans;
    }
    public char[][] ndc(int N) {
    	char[][] ans = new char[N][];
    	for (int i = 0; i < N; i++) {
			ans[i] = nc();
		}
    	return ans;
    }
    public int[][] nddi(int N, int S) {
    	int[][] ans = new int[N][S];
    	for (int i = 0; i < N; i++) {
			for (int j = 0; j < S; j++) {
				ans[i][j] = ni();
			}
		}
    	return ans;
    }
}