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 <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cctype>
#include <climits>
#include <string>
#include <bitset>
using namespace std;
typedef long double ld;
typedef long long int ll;
typedef unsigned long long int ull;
typedef vector<int> vi;
typedef vector<char> vc;
typedef vector<double> vd;
typedef vector<string> vs;
typedef vector<ll> vll;
typedef pair<int,int> P;
typedef vector<pair<int,int> > vpii;
typedef vector<pair<string,int>> vpsi;
typedef vector<vector<int> > vvi;
typedef vector<vector<char> > vvc;
typedef vector<vector<string> > vvs;
typedef vector<vector<ll> > vvll;
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define fin(ans) cout<<(ans)<<endl
#define STI(s) atoi(s.c_str())
#define mp(p,q) make_pair(p,q)
#define pb(n) push_back(n)
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define Sort(a) sort(a.begin(),a.end())
#define Rort(a) sort(a.rbegin(),a.rend())
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;}
const int MOD = 1000000007;
const int INF = INT_MAX;
const ll LLINF = 1LL<<60;


P f(int x){
    int a=x%10;
    int b=0;
    while(x){
        b=x;
        x/=10;
    }
    return P(a,b);
}

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

int n;
cin >> n;
map<P,int> m;
for(int i=1;i<=n;++i){
    m[f(i)]++;
}
ll ans=0;
for(int i=1;i<=n;++i){
    P p = f(i);
    P q(p.second,p.first);
    ans+=m[q];
}
fin(ans);

//////////////////////////////////////////////////////
return 0;}