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 fi first
#define se second
#define countof(array) (sizeof(array) / sizeof(array[0]))
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = (n)-1; i >= 0; --i)
#define rep2(i,n) for(int i = 1; i <= (n); ++i)
#define rrep2(i,n) for(int i = (n); i > 0; --i)
#define srep(i,s,n) for(int i = s; i < (n); ++i)
#define rsrep(i,s,n) for(int i = (n)-1; i >= s; --i)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define aall(a) (a), (a)+countof(a)//for array sorting
#define raall(a) (a), (a)+countof(a), greater<>()
#define show(x) cout<<#x<<" = "<<(x)<<endl;
#define vfind(v, a) find(all(v), a) != v.end()
#define yn(f) { if (f) puts("YES"); else puts("NO"); }
#define yns(f) { if (f) puts("Yes"); else puts("No"); }
#define show_ary(...) { cout<<#__VA_ARGS__<<" = "; for (const auto& x : (__VA_ARGS__)) { cout<<x<<" "; } cout<<endl; }
#define show_pair(...) cout<<#__VA_ARGS__<<" = "<<endl; for (const auto& x : (__VA_ARGS__)) { cout<<"  "<<x.fi<<" : "<<x.se<<endl; }
#define out_ary(...) { int n = (__VA_ARGS__).size(); rep(i, n) printf("%d%s", (__VA_ARGS__)[i], i != n-1 ? " " : "\n"); }
#define argmax(v) distance((v).begin(), max_element(all(v)))
#define argmin(v) distance((v).begin(), min_element(all(v)))
#define vmax(v) *max_element(all(v))
#define vmin(v) *min_element(all(v))
typedef long long int ll;
typedef pair<int, int> P;
typedef vector<P> vpair;
typedef vector<int> vint;
typedef vector<ll> vll;
typedef vector<double> vdouble;
typedef vector<string> vstr;
typedef vector<bool> vbool;
typedef vector<vint> vvint;
typedef vector<vll> vvll;
typedef vector<vstr> vvstr;
typedef vector<vbool> vvbool;
const ll LINF = 2000000000000000000ll;
const int INF = 1000000100;
const ll MOD = 1e9+7;


int main() {
    int n;
    cin >> n;

    int ans = 0;
    vvint c(10, vint(10, 0));
    rep2(i, n) {
        int a = i%10;
        int b; 
        int tmp = i;
        while(tmp!=0) {
            if (tmp < 10) {
                b = tmp;
                break;
            }
            tmp /= 10;
        }

        if (a == 0 || b == 0) continue;

        if (c[b][a] != 0) ans += c[b][a];
        else {
            if (a == b && a <= n) c[b][a]++;

            rep(i, 5) {
                int num = a;
                int mul = pow(10, i);
                rep(j, mul) {
                    int tmp = (num*mul*10 + j*10) + b;
                    if (tmp <= n) c[b][a]++;
                }
            }

            ans += c[b][a];
        }
    }
    cout << ans << endl;
    return 0;
}