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)

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    ll N; cin >> N;
    ll ans = 0;

    map<P,int> cnt;

    for (int i = 1;i < N;i++){
        string s = to_string(i);
        cnt[P(s[0]-'0',s[s.size()-1]-'0')]++;
    }

    for (auto&& i:cnt){
        ans += i.second*cnt[P(i.first.second,i.first.first)];
    }

    cout << ans << endl;

    return 0;

}