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: Wrong Answer)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 

using namespace __gnu_pbds; 
using namespace std;

#define int           long long 
#define S             second
#define F             first
#define pb            push_back
#define all(c)        (c).begin(),(c).end()
#define rall(c)       (c).rbegin(),(c).rend() 
#define lb            lower_bound
#define ub            upper_bound
#define si(c)         (int)((c).size())
#define lcm(a, b)      (a * (b / __gcd(a,b)))  
#define inf           (int)(4e18)
// #define endl          '\n'
#define mp            make_pair
#define time(s)       (double(clock()-s)/double(CLOCKS_PER_SEC))
#define debug(args...)    _F(#args, args)
#define vi            std::vector<int>
#define pii           pair<int, int> 
#define vpi           vector<pii>
#define ordered_set   tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update> 

clock_t start;
mt19937_64 rng(chrono::system_clock::now().time_since_epoch().count());

template<typename T> void _F(const char *name, T arg1){ cerr << name << " = " << arg1 << endl;}
template<typename T, typename... Args> void _F(const char *names, T arg1, Args... args)
{ const char *name = strchr(names, ',');cerr.write(names, name-names) << " = " << arg1 << endl;_F(name+2, args...);}

template< typename T1, typename T2 > istream& operator>>(istream& in, pair<T1, T2> &q){ in >> q.F >> q.S; return in;}
template< typename T1, typename T2 > ostream& operator<<(ostream& out, pair<T1, T2> &q){ out << q.F << " " << q.S; return out;}
template< typename T1, typename T2 > pair<T1, T2> operator+(pair<T1, T2> p1, pair<T1, T2> p2){ return {p1.F+p2.F, p1.S+p2.S};}
template< typename T1, typename T2 > pair<T1, T2> operator-(pair<T1, T2> p1, pair<T1, T2> p2){ return {p1.F-p2.F, p1.S-p2.S};}
template< typename T1, typename T2 > bool operator<(pair<T1, T2> p1, pair<T1, T2> p2){ return p1 < p2 ;}

template<typename T> void Unique(vector<T> &v){
    sort(all(v)), v.resize(distance(v.begin(), unique(all(v))));
}

const int M = 1e9+7;
const int N = 1e6+1;

int f[N], inv[N];

int Pow(int x, int p) {
	int ans = 1;
	while(p) {
		if(p&1) ans = (ans*x)%M;
		x = (x*x)%M;
		p >>= 1;
	}
	return ans;
}

int ncr(int n, int r) {
	return f[n]*inv[r]%M*inv[n-r]%M;
}

void pre() {
	f[0] = inv[0] = 1;
	for(int i = 1; i < N; i++) {
		f[i] = f[i-1]*i%M;
		inv[i] = Pow(f[i], M-2);
	}
}

void solve() {
	pre();
	int n; cin >> n;
	int ans = 0;
	for(int i = 2; i <= n; i++) {
		int add = ncr(n, i);
		add *= (Pow(2, i)-2+M)%M;
		ans += add;
		ans %= M;
	}
	cout << ans << endl;
}

signed main(){

    ios_base::sync_with_stdio(0); 
    cin.tie(0); cout.tie(0);   
    start = clock(); 

    // #ifndef ONLINE_JUDGE
    // freopen("input.txt", "r", stdin); 
    // freopen("output.txt", "w", stdout);
    // freopen("error.txt", "w", stderr);
    // #endif

    int test = 1;
    // cin >> test;
    
    cout << fixed << setprecision(2);
     
    for(int i = 1; i <= test; ++i){
    	solve();
    }

    cerr << time(start);
    return 0;
}