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>
using namespace std;
#define mid ((left + right) >> 1)
#define lson rt << 1, left, mid
#define rson rt << 1 | 1, mid + 1, right

const double pi = acos(-1.0);
const int INF = 1e9 + 1;
const int N = 1e5 + 5;

int n;
int sum[11][11];

int main() {
    ios::sync_with_stdio(false);
    cin >> n;
    for (int i = 1; i <= n; i++) {
        int val = i;
        int head = -1;
        while (val >= 10) val /= 10;
        head = val;
        int tail = i % 10;
        sum[head][tail] += 1;
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
        int val = i;
        int head = -1;
        while (val >= 10) val /= 10;
        head = val;
        int tail = i % 10;
        if (tail == 0) continue;
        ans += sum[tail][head];
    }
    cout << ans << endl;
    return 0;
}