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)

// ABC 152-D
// 2020.1.19 bal4u

#include <stdio.h>

#define gc() getchar_unlocked()

int in() {   // 非負整数の入力
	int n = 0, c = gc();
	do n = 10 * n + (c & 0xf); while ((c = gc()) >= '0');
	return n;
}

int f1[100], f2[100];

void calc(int n) {
	int k = n % 10;
	while (n >= 10) n /= 10;
	++f1[10*k + n], ++f2[n*10 + k];
}

int main()
{
	int i, N, ans;
	
	N = in();
	while (N) calc(N--);

	ans = 0;
	for (i = 0; i < 100; ++i) ans += f1[i] * f2[i];
	printf("%d\n", ans);
	return 0;
}