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 <algorithm>
#include <utility>
#include <vector>

template <class T, class U>
inline bool chmin(T& lhs, const U& rhs) {
  if (lhs > rhs) {
    lhs = rhs;
    return true;
  }
  return false;
}

template <class T, class U>
inline bool chmax(T& lhs, const U& rhs) {
  if (lhs < rhs) {
    lhs = rhs;
    return true;
  }
  return false;
}

// [l, r) from l to r
struct frep {
  struct itr {
    int i;
    constexpr itr(int i_): i(i_) { }
    constexpr void operator ++ () { ++i; }
    constexpr int operator * () const { return i; }
    constexpr bool operator != (itr x) const { return i != x.i; }
  };
  const itr l, r;
  constexpr frep(int l_, int r_): l(std::min(l_, r_)), r(r_) {}
  constexpr itr begin() const { return l; }
  constexpr itr end() const { return r; }
};

// [l, r) from r to l
struct brep {
  struct itr {
    int i;
    constexpr itr(int i_): i(i_) { }
    constexpr void operator ++ () { --i; }
    constexpr int operator * () const { return i; }
    constexpr bool operator != (itr x) const { return i != x.i; }
  };
  const itr r, l;
  constexpr brep(int r_, int l_): r(std::max(l_, r_) - 1), l(l_ - 1) {}
  constexpr itr begin() const { return r; }
  constexpr itr end() const { return l; }
};

template <class T>
inline T scan() {
  T res;
  std::cin >> res;
  return res;
}

int calc(int x) {
  int res = 1;
  while (x >= 10) {
    x /= 10;
    res *= 10;
  }
  return res;
}

int main() {
  const int N = scan<int>();
  long long count[10][10] = {};
  for (int i: frep(1, N + 1)) {
    count[i / calc(i)][i % 10]++;
  }
  long long ans = 0;
  for (int i: frep(1, 10)) {
    for (int j: frep(1, 10)) {
      ans += count[i][j] * count[j][i];
    }
  }
  std::cout << ans << '\n';
  return 0;
}