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 rep(i,n) for(int i=0;i<n;i++)
#define REP(i,s,n) for(int i=(s);i<=(n);i++)
#define repr(i,n) for(int i=n-1;i>=0;i--)
#define REPR(i,s,n) for(int i=(s);i>=(n);i--)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define Eunique(v) v.erase(unique(all(v)),v.end())
#define Eback(s) s.erase(s.end()-1,s.end())
#define rev(v) reverse(all(v))
#define minvec(v) *min_element(all(v))
#define maxvec(v) *max_element(all(v))
#define sumvec(v) accumulate(all(v),0)
#define mapmax(v) v.rbegin()->first
#define mapmin(v) v.begin()->first
#define pb push_back
#define pf push_front
#define m_p make_pair
#define DOUBLE fixed << setprecision(15)
#define OK puts("OK")
#define OK1 puts("OK1")
#define OK2 puts("OK2")
#define SIZE(s) (int)s.size()
#define INF ((1LL<<62)-(1LL<<31))
#define zero(x,n) setw(x) << setfill('0') << n

typedef long long ll;

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<long long> vll;
typedef vector<vll> vvll;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<char> vc;
typedef vector<vc> vvc;
typedef vector<bool> vb;
typedef vector<vb> vvb;
typedef vector<string> vs;

typedef pair<ll,ll> pll;
typedef pair<int,int> pii;
typedef vector<pair<int,int>> vpii;
typedef vector<pair<ll,ll>> vpll;

const double pi = acos(-1.0);
const int mod=1000000007;
const int mod2=998244353;

template<class A, class B>
ostream& operator<<(ostream& ost, const pair<A, B>&p) {
	ost << "{" << p.first << ", " << p.second << "} ";
	return ost;
}

template<class T>
ostream& operator<<(ostream& ost, const vector<T>&v) {
	ost << "{";
	for (int i = 0; i<(int)v.size(); i++) {
		if (i)ost << " ";
		ost << v[i];
	}
	ost << "} \n";
	return ost;
}

template<class A, class B>
ostream& operator<<(ostream& ost, const map<A, B>&v) {
	ost << "{";
	for (auto p:v) {
		ost << "{" << p.first << ", " << p.second << "} ";
	}
	ost << "} ";
	return ost;
}

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

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

void YES(bool b) {cout << (b?"YES\n":"NO\n");}
void Yes(bool b) {cout << (b?"Yes\n":"No\n");}
void yes(bool b) {cout << (b?"yes\n":"no\n");}
void Yay(bool b) {cout << (b?"Yay!\n":":(\n");}

long long powmod(long long a,long long b){
    long long c=1;
    while(b>0){
        if(b&1) c=a*c%mod;
        a=a*a%mod;
        b>>=1;
    }
    return c;
}

ll gcd(ll x, ll y) {return __gcd(x,y);}
ll lcm(ll x, ll y) {return x/__gcd(x,y)*y;}

vector<long long> divisor(long long n){
    vector<long long> ret;
    for(int64_t i=1;i*i<=n;i++){
        if(n%i!=0) continue;
		ret.push_back(i);
		if(i*i!=n) ret.push_back(n/i);
    }
    sort(ret.begin(),ret.end());
    return ret;
}

int main()
{
	ll n,ans=0;
	vvll c(10,vll(10));
	cin >> n;
	for(ll i=1;i<=n;i++){
		string s=to_string(i);
		ll m=(ll)s.size();
		c[s[0]-'0'][s[m-1]-'0']++;
	}
	REP(i,1,9){
		REP(j,1,9){
			ans+=c[i][j]*c[j][i];
		}
	}
	cout << ans << endl;
	return 0;
}