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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#pragma GCC optimize("O3")
#pragma GCC optimize ("unroll-loops")
#pragma GCC target ("avx")
#include <bits/stdc++.h>
constexpr int INF = 2147483647;
constexpr long long int INF_LL = 9223372036854775807;
constexpr int MOD = 1000000007;
constexpr double PI = 3.14159265358979323846;
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;


// AxxxB
// BxxxA
int main() {
	int N;
	cin >> N;
	int ans = 0;
	map<pair<int, int>,int> p;
	for (int i = 1; i < N; i++) {
		auto s = to_string(i);
		p[{s[0] - '0', s.back() - '0'}]++;
	}
	for (int i = 0; i < 10; i++) {
		for (int j = 0; j < 10; j++) {
			ans += p[{i, j}] * p[{j, i}];
		}
	}
	cout << ans << endl;
}