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 <vector>
#include <stack>
#include <queue>
#include <list>
#include <bitset>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <algorithm>
#include <numeric>
#include <iostream>
#include <iomanip>
#include <string>
#include <chrono>
#include <random>
#include <cmath>
#include <cassert>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <functional>
#include <sstream>

using namespace std;

class Solution {
public:
    long long solve(int n) {
        vector<vector<int>> cnts(10, vector<int>(10, 0));

        for (int i = 1; i <= n; ++i) {
            int x = i;
            int last = x % 10;
            int first = last;
            while (x != 0) {
                first = x % 10;
                x /= 10;
            }
            ++cnts[first][last];
        }

        long long res = 0;
        for (int i = 0; i < 10; ++i) {
            for (int j = 0; j < 10; ++j) {
                res += cnts[i][j] * 1LL * cnts[j][i];
            }
        }
        return res;
    }
};

int main(int argc, char** argv) {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n;
    cin >> n;

    Solution sol;
    cout << sol.solve(n) << "\n";

    return 0;
}