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>
#define del(a,i) memset(a,i,sizeof(a))
#define ll long long
#define inl inline
#define il inl void
#define it inl int
#define ill inl ll
#define re register
#define ri re int
#define rl re ll
#define INF 0x3f3f3f3f
#define lowbit(x) (x&(-x))
#define mid ((l+r)>>1)
using namespace std;
template<class T>il read(T &x){
	x=0;int f=1;char k=getchar();
	for(;k>'9'||k<'0';k=getchar()) if(k=='-') f=-1;
	for(;k>='0'&&k<='9';k=getchar()) x=x*10+k-'0';
	x*=f;
}
template<class T>il _print(T x){
	if(x>9) _print(x/10);
	putchar(x%10+'0');
}
template<class T>il print(T x){
	if(x<0) putchar('-'),x=-x;
	_print(x);
}
int n,cnt[20][20];
ll ans;
int main(){
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	read(n);
	for(ri i=1;i<=n;++i){
		int pre=i,lst=i%10;
		while(pre>=10) pre/=10;
		ans+=2*cnt[lst][pre];
		++cnt[pre][lst];
		if(pre==lst) ++ans;
	}
	print(ans);
	return 0;
}