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;
typedef long long ll;
using pint = pair<int,int>;
ll mod = 1000000007,mod2 = 998244353;

void chmin(long long &a, long long b) { if (a > b) a = b; }
void chmax(long long &a, long long b) { if (a < b) a = b; }
const long long INF = 1LL<<60;
ll gcd(ll a,ll b){
    if(b == 0) return a;
    else return gcd(b,a%b);
}

ll lcm(ll a,ll b){
    ll g = gcd(a,b);
    a /= g;
    return a*b;
}

//重みある時
using Edge = pair<int,int>;
using Graph = vector<vector<Edge> >;

int main() {
    ll n;
    cin >> n;
    vector<vector<int>> c(10,vector<int>(10,0));
    for (int i = 1; i <= n; ++i) {
        vector<int> dig;
        int tmp = i;
        while(tmp > 0){
            dig.push_back(tmp%10);
            tmp /= 10;
        }
        c[dig[dig.size()-1]][dig[0]]++;
    }
    ll ans = 0;
    for (int i = 1; i <= 9; ++i) {
        for (int j = 1; j <= 9; ++j) {
            ans += c[i][j]*c[j][i];
        }
    }
    cout << ans << endl;
    return 0;
}