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: Accepted)

#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<(n);++i)
#define rrep(i,n) for(int i=1;i<(n);++i)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define maxs(a, b) a = max(a, b)
#define mins(a, b) a = min(a, b)
using namespace std;
typedef long long ll;
typedef pair<double, int> P;
const ll linf = (1ll << 61);
const int inf = 1001001001;
const int mod = 1000000007;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n;
	cin >> n;
	vector<vector<int>> cnt(10, vector<int>(10));
	for (int i = 1; i <= n; ++i) {
		int a = i % 10;
		int b = i;
		while (b >= 10) b /= 10;
		cnt[a][b]++;
	}
	int ans = 0;
	rep(i, 10)rep(j, 10) ans += cnt[i][j] * cnt[j][i];
	cout << ans << endl;
 	return 0;
}