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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <iomanip>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <cstdio>
#include <utility>
#include <string>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <deque>
#include <numeric>

using namespace std;

typedef uint64_t u64;
typedef int64_t s64;
typedef uint32_t u32;
typedef int32_t s32;
typedef vector<s32> vs32;
typedef vector<u32> vu32;
typedef vector<s64> vs64;
typedef vector<u64> vu64;

const double PI=3.14159265358979323846;

#define MAX(x, y) ((x) < (y) ? (y) : (x))
#define MIN(x, y) ((x) > (y) ? (y) : (x))

#define rep(i, N) for(int i = 0; i < N; ++i)

#define CEIL(x, y) (((x) + (y) - 1) / (y))
#define MOD 1000000007ULL

#define IN(l, r, x) ((l) <= (x) && (x) < (r))

int cnt[100];

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n;
    cin >> n;

    rep (i, n + 1)
    {
        int u = i;
        while (u >= 10) u /= 10;
        ++cnt[u * 10 + i % 10];
    }

    int ans = 0;
    for (int i = 1; i < 10; ++i)
    {
        for (int j = 1; j < 10; ++j)
        {
            ans += cnt[i * 10 + j] * cnt[j * 10 + i];
        }
    }

    cout << ans << "\n";
    return 0;
}