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>
const int INF = 1e9;
const int MOD = 1e9+7;
const long long LINF = 1e18;
#define dump(x)  cout << 'x' << ' = ' << (x) << ` `;
#define FOR(i,a,b) for(ll i=(a);i<(b);++i)
#define rep(i,n) for(ll i=0;i<(n);++i)
#define REPR(i,n) for(ll i=n;i>=0;i--)
#define FOREACH(x,a) for(auto& (x) : (a) )
typedef long long ll;
using namespace std;
typedef pair<ll, ll> P;

template<typename T>
void print(const vector<T> &x) {
    int n = x.size();
    rep(i,n) {
        cout << x[i];
        if (i!=n-1) cout<<" ";
        else cout << endl;
    }
}

template<typename T>
void print(const vector<vector<T>> &x) {
    int n = x.size();
    rep(i,n) {
        rep(j,x[i].size()) {
            cout << x[i][j] << " ";
        }
        cout << endl;
    }
}

template<typename T>
void print(const vector<T> &x, int n) {
    rep(i,n) {
        cout << x[i];
        if (i!=n-1) cout<<" ";
        else cout << endl;
    }
}

template<typename T>
void print(const vector<vector<T>> &x, int n, int m) {
    rep(i,n) {
        rep(j,m) {
            cout << x[i][j] << " ";
        }
        cout << endl;
    }
}

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

void input_init() {
    cin.tie(0); ios::sync_with_stdio(false);
}

int main(int argc, char const *argv[]) {
    ll n; cin>>n;
    ll cnt[10][10];
    memset(cnt,0,sizeof(cnt));

    for (int i = 1; i <= n; ++i) {
        string s = to_string(i);
        cnt[s.front()-'0'][s.back()-'0']++;
    }

    ll ans = 0;
    for (int i = 1; i <= n; ++i) {
        string s = to_string(i);
        ans += cnt[s.back()-'0'][s.front()-'0'];
    }

    cout << ans << '\n';
    return 0;
}