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 debug(x) cout <<"DEBUG"<<" "<< #x << ":" << x << '\n'

//↓0-originか1-originでn回繰り返し
#define rep(i,n) for(int i = 0; i < ((int)(n)); i++)   //0-indexed昇順
#define rep1(i,n) for(int i = 1; i <= ((int)(n)); i++) //1-indexed昇順
#define rrep(i,n) for(int i = ((int)(n)-1); i >= 0; i--)  //0-indexed降順
#define rrep1(i,n) for(int i = ((int)(n)); i >= 1; i--)   //1-indexed降順
#define all(x) (x).begin(),(x).end()

typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<vi> vii; //2次元配列
typedef vector<ll> vll;
typedef pair<int, int> PII;

template<typename T>
using vec = vector<T>;
template<class T>
using maxheap = std::priority_queue<T>;
template<class T>
using minheap = std::priority_queue<T, std::vector<T>, std::greater<T>>;

template<class T,class U> inline bool chmax(T &a, const U &b){if(a<b){a=b;return 1;}return 0;}
template<class T,class U> inline bool chmin(T &a, const U &b){if(a>b){a=b;return 1;}return 0;}

const ld PI = acos(-1.0L);
const ll infll = (1LL<<62)-1;
const int inf = (1<<30)-1;
const int mod = 1000000007;
//const int mod = 998244353;


/*FUNCs=================================================*/


/*MAIN==================================================*/
signed main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);   //cin cout 高速化
    //cout << fixed << setprecision(10);

    int n; cin >> n;
    int cnt = 0;
    vii p(10, vi(10));
    string s;
    for(int i = 1; i <= n; ++i) {
        s = to_string(i);
        int x = s[0]-'0';
        int y = s[s.size()-1]-'0';
        p[x][y]++;
    }
    rep(i,10) {
        rep(j,10) {
            cnt += p[i][j] * p[j][i];
        }
    }
    cout << cnt << '\n';
}