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>
#include <boost/multiprecision/cpp_int.hpp>

/***** type *****/
using namespace std;
using ll = long long;
using ld = long double;
template <class T> using vt = vector<T>;
template <class T> using vvt = vector<vector<T>>;
template <class T> using vvvt = vector<vector<vector<T>>>;
template <class T1, class T2> using umap = unordered_map<T1, T2>;
using ml = boost::multiprecision::cpp_int;

/***** define *****/
#define all(c) (c).begin(), (c).end()            // begin to end
#define rep(i, b, e) for (ll i = b; i < e; i++)  // repeat
#define repr(i, b, e) for (ll i = b; e < i; i--) // repeat reverse
#define val(itr) *(itr)                          // get value
/***** const value *****/
#define llong_max 9223372036854775807            // 9 * 10^18
#define ldbl_max 1.79769e+308                    // 1.7 * 10^308
#define pi 3.1415926535897932                    // 3.14 ...
/***** for each macro *****/
#define fori(i, ...) if(ll i = -1) for(__VA_ARGS__) if(i++, true)
#define each(i, e, c) fori(i, auto e = c.begin(); e != c.end(); ++e)
#define forir(i, v, ...) if(ll i=(ll)v.size())for(__VA_ARGS__)if(i--,true)
#define eachr(i, e, c) forir(i, auto e = c.rbegin(); e != c.rend(); ++e)

/***** lambda *****/
auto Count = [] // long long count value
(auto b, auto e, auto x) { return (ll)count(b, e, x); };
auto CtoN = [] // char to number
(auto c) { return (ll)(c - '0'); };
auto DivC = [] // long double div ceiling
(auto a, auto b) { return ceil((ld)a / (ld)b); };
auto Fix = [] // fix value
(auto b, auto e, auto fix)
{ for (auto it = b; it != e; ++it)*it += fix; };
auto Pow = [] // long long pow
(auto a, auto b) { return (ll)pow(a, b); };
auto Pow2 = [] // long long pow2
(auto n) { return (1LL << n); };
auto Pow10 = [] // long long pow10
(auto n) { return (ll)pow(10, n); };
auto Size = []  // long long collection size
(auto& c) { return (ll)(c).size(); };
auto Sum = [] // long long accumulate
(auto b, auto e) { return accumulate(b, e, 0LL); };

/***** template *****/
template <class T> vvt<T> VVT
(ll ys, ll xs, T fill = T())
{	// vector<vector<T>> resize + fill
	vvt<T> v(ys);
	each(i, y, v) { y->resize(xs, fill); } return v;
}
template <class T> vvvt<T> VVVT
(ll zs, ll ys, ll xs, T fill = T())
{	// vector<vector<vector<T>>> resize + fill
	vvvt<T> v(zs);
	each(i, z, v) { val(z) = VVT(ys, xs, fill); } return v;
}
template <class T> vt<T> InputVT
(ll size, T fix = T())
{	// input vector<T> (T != struct) + fix
	vt<T> v(size);
	rep(i, 0, size) { cin >> v[i]; v[i] += fix; }
	return v;
}
template <class T> vvt<T> InputVVT
(ll ys, ll xs, T fix = T())
{	// input vector<vector<T>> (T != struct) + fix
	vvt<T> v(ys); each(i, y, v) y->resize(xs);
	rep(y, 0, ys) rep(x, 0, xs) { cin >> v[y][x]; v[y][x] += fix; }
	return v;
}
template <class T> vvvt<T> InputVVVT
(ll zs, ll ys, ll xs, T fix = T())
{	// input vector<vector<vector<T>>> (T != struct) + fix
	vvvt<T> v(zs);
	each(i, z, v) { val(z) = InputVVT<T>(ys, xs, fix); }
	return v;
}

//using namespace NyaGadget;
//using mll = ModLL<1000000007>;

struct NyaNum
{
	ll s = 0;
	ll e = 0;
};

bool operator< (const NyaNum& l, const NyaNum& r)
{	// priority_queueを降順でソートする
	// 昇順にしたいときは不等号を「>」にする
	if (l.s != r.s) return l.s < r.s;
	return l.e < r.e;
};

NyaNum GetNyaNum(ll x)
{
	NyaNum nn;
	nn.e = x % 10;
	while (x)
	{
		nn.s = x;
		x /= 10;
	}
	return nn;
}

int main(void)
{
	ll N; cin >> N;

	map<NyaNum, ll> m;
	rep(i, 1, N+1)
	{
		NyaNum nn = GetNyaNum(i);
		m[nn]++;
	}

	ll ans = 0;
	rep(i, 1, N+1)
	{
		NyaNum nn = GetNyaNum(i);
		ans += m[{nn.e, nn.s}];
	}
	cout << ans;
	return 0;
}