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 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define IOS ios::sync_with_stdio(false); cin.tie(0);
#define FOR(i,s,n) for(int i = (s); i < (n); i++)
#define REP(i,n) FOR(i,0,n)
#define RREP(i,n) for(int i = (n); i >= 0; i--)
#define ALL(n) (n).begin(), (n).end()
#define RALL(n) (n).rbegin(), (n).rend()
#define ATYN(n) cout << ( (n) ? "Yes":"No") << '\n';
#define CFYN(n) cout << ( (n) ? "YES":"NO") << '\n';
#define OUT(n) cout << (n) << '\n';
using ll = long long;
using ull = unsigned long long;
using pii = pair<int,int>;
using pll = pair<ll,ll>;

int main(void)
{
    IOS
    int N; cin >> N;
    map<pair<char,char>,int> mp;
    REP(i,N+1) {
        string s = to_string(i);
        mp[{s.front(),s.back()}]++;
    }

    int ans = 0;
    FOR(i,'1','9'+1) FOR(j,'1','9'+1) ans += mp[{i,j}] * mp[{j,i}];
    cout << ans << '\n';

    return 0;
}