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 C (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<limits.h>

#define rep(i,begin,end) for(int i=begin; i<end; i++)
#define revrep(i,begin,end) for(int i=begin; i>end; i--)
#define lld long long int

int main(){
    lld n;
    scanf("%lld", &n);
    lld t[10][10];
    rep(i, 0, 10){
        rep(j, 0, 10){
            t[i][j] = 0;
        }
    }
    rep(i, 1, n+1){
        lld buf = i;
        lld a, b;
        b = buf % 10;
        while(buf >= 10){
            buf /= 10;
        }
        a = buf;
        t[a][b]++;
    }
    lld ans = 0;
    rep(i, 1, 10){
        rep(j, 1, 10){
            ans += t[i][j] * t[j][i];
            //printf("%lld ", t[i][j]);
        }
        //printf("\n");
    }
    printf("%lld", ans);

    return 0;
}