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 < (int)(n); ++i)
#define reps(i,n) for (int i = 1; i <= (int)(n); ++i)
#define repd(i,n) for (int i = (int)(n - 1); i >= 0; --i)
#define repds(i,n) for (int i = (int)(n); i > 0; --i)
#define loop(i,x,n) for (int i = (int)(x); i < (n); ++i)
#define loops(i,x,n) for (int i = (int)(x); i <= (n); ++i)
#define loopd(i,x,n) for (int i = (int)(x); i > (n); --i)
#define loopds(i,x,n) for (int i = (int)(x); i >= (n); --i)
#define itrep(i,s) for (auto i = begin(s); i != end(s); ++i)
#define itrepd(i,s) for (auto i = --end(s); i != begin(s); --i)
#define all(f,x,...) [&](decltype(x)& whole) { return (f)(begin(whole), end(whole), ##__VA_ARGS__); }(x)
#define rall(f,x,...) [&](decltype(x)& whole) { return (f)(rbegin(whole), rend(whole), ##__VA_ARGS__); }(x)
using ll = long long;
using ld = long double;
constexpr ll inf = static_cast<ll>(1e17);
constexpr int iinf = static_cast<int>(1e9);
constexpr double dinf = 1e10;
constexpr ld ldinf = 1e17;
ll mod = static_cast<ll>(1e9 + 7);
std::ostream& endn(std::ostream& os) { return os.put(os.widen('\n')); }
template<class T> constexpr int sz(const T& a) { return (int)a.size(); }
template<class T> constexpr T pow2(const T a) { return a * a; }
template<class T, class ... Args> constexpr T mins(T& a, Args... args) { return a = min<T>({ a, args... }); }
template<class T, class... Args> constexpr T maxs(T& a, Args... args) { return a = max<T>({ a,args... }); }
template<typename A, size_t N, typename T>
void fillArray(A(&array)[N], const T& val) { std::fill((T*)array, (T*)(array + N), val); }

int n;
void solve() {
    cin >> n;

    int c = 0;
    int tmp = n;
    while (tmp != 0) {
        ++c;
        tmp /= 10;
    }

    ll ans = 0;
    reps(l, 9) reps(r, 9) {
        ll res1 = 0, res2 = 0;
        rep(i, c) {
            int e = (l + 1) * pow(10, i);
            loop(j, l * pow(10, i), e) {
                if (j > n) break;
                if (j % 10 != r) continue;
                ++res1;
            }
            int e2 = (r + 1) * pow(10, i);
            loop(j, r * pow(10, i), e2) {
                if (j > n) break;
                if (j % 10 != l) continue;
                ++res2;
            }
        }
        ans += res1 * res2;
    }

    cout << ans << endn;

}

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    solve();
    cout << flush;
    return 0;
}