Problem:
Score : 300 points

Problem Statement

How many integer sequences A_1,A_2,\ldots,A_N of length N satisfy all of the following conditions?

0 \leq A_i \leq 9

There exists some i such that A_i=0 holds.

There exists some i such that A_i=9 holds.

The answer can be very large, so output it modulo 10^9 + 7.

Constraints

1 \leq N \leq 10^6

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the answer modulo 10^9 + 7.

Sample Input 1

2

Sample Output 1

2

Two sequences \{0,9\} and \{9,0\} satisfy all conditions.

Sample Input 2

1

Sample Output 2

0

Sample Input 3

869121

Sample Output 3

2511445

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

// in case of multiple test cases and N=1e5 don't intialize arrays as global except for graph vector-array
// always check whether or not you are doing mod of a negative number
// always use 1LL instead of 1
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
 
using namespace __gnu_pbds;
using namespace std;

// using long doubles saves you from corner cases but is very time consuming
#define double      	long double
#define int         	long long
#define pb          	push_back
#define pii         	pair<int,int>
#define vi          	vector<int>
#define vii         	vector<pii>
#define mi          	map<int,int>
#define mii         	map<pii,int>
#define all(a)      	(a).begin(),(a).end()
#define sz(x)       	(int)x.size()
// just comment the line below in case of interactive problems
#define endl        	"\n"
#define repp(i,a,b) 	for(int i=a;i<b;i++)
#define rep(i,a,b) 		for(int i=a;i<=b;i++)
#define brep(i,a,b)     for(int i=a;i>=b;i--)
#define deb1(x)      	cout << #x << "=" << x << endl
#define deb2(x, y)  	cout << #x << "=" << x << "," << #y << "=" << y << endl
#define deb3(x, y, z)  	cout << #x << "=" << x << "," << #y << "=" << y << "," << #z << "=" << z << endl
#define trace(v) 		for(auto it=v.begin();it!=v.end();it++)cout<<*it<<" ";cout<<endl;
#define tracearr(a,l,r) for(int iii=l;iii<=r;iii++)cout << a[iii] << " ";cout << endl;
#define PI          	3.1415926535897932384626
#define F 				first
#define S 				second
#define clr(x,y) 		memset(x, y, sizeof(x))
#define IOS         	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);


typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
// *s.find_by_order(1) - gives the 2nd samllest element in set
// s.order_of_key(x)   - gives the number of elements in the set which are strictly smaller than x


const int N=1e6 + 5;
const int MOD=1e9 + 7;

vector<long long> factorial(N, 1);// initialized with 1
 
inline long long add(long long x, long long y) {
	return ((x % MOD) + (y % MOD))% MOD;
}
 
inline long long subtract(long long x, long long y) {
	x = x % MOD;
	y = y % MOD;
 
	if(x >= y) return x - y;
	else return x - y + MOD;
}
 
inline long long multiply(long long x, long long y) {
	return ((x % MOD) * (y % MOD)) % MOD;
}
 
inline long long power(long long x, long long y) {
	long long res = 1;
	x = x % MOD;
 
	while(y > 0) {
		if(y & 1) {
			res = multiply(res, x);
		}
		y = y / 2;
		x = multiply(x, x);
	}
 
	return res;
}
 
inline long long divide(long long x, long long y) {
	return multiply(x, power(y, MOD-2));
}
 
inline long long nCr(long long n, long long r) {
	if(n < r or r < 0 or n < 0) return 0;
	return divide(factorial[n], multiply(factorial[r], factorial[n-r]));
}

// in nCr, if n is huge like 1e9 we can use this instead of pre computing the factorial method
inline long long nCm(long long x,long long y){
    if (x < y)
        return 0;
    long long res = 1;
    for (long long i = 0; i < y; i++) { 
        res *= (x - i);res%=MOD; 
        res *= power(i + 1,MOD-2);
        res%=MOD; 
    }
    return res;
}
 
void precompute() {
	for(int i = 1; i < N-1; i++) {
		factorial[i] = multiply(factorial[i-1], i);
	}
}

int n;
int dp[N][2][2];

int solve(int i,int c1,int c2)
{
	if(i>n)
	{
		if(c1==1 && c2==1)
			return 1LL;
		else
			return 0LL;
	}
	int& ans = dp[i][c1][c2];
	if(ans!=-1)
		return ans;
	ans = 0;
	rep(j,0,9)
	{
		if(j==0)
			ans = add(ans,solve(i+1,1,c2));
		else if(j==9)
			ans = add(ans,solve(i+1,c1,1));
		else
			ans = add(ans,solve(i+1,c1,c2));
	}
	return ans;
}

int32_t main()
{
	IOS;
	int T=1;
	//cin >> T;
	// int numOfSetBits = __builtin_popcountll(n);
	std::cout << std::fixed << std::setprecision(12);
	while(T--)
	{
		cin >> n;
		if(n==1)
		{
			cout << "0";
			return 0;
		}
		clr(dp,-1);
		cout << solve(1,0,0);
	}
}