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>
using namespace std;

#define print(a)        for (auto x : a) cout << x << " "; cout << endl
#define print_upto(a,n)        for(int i=0;i<n;i++)    cout<<a[i]<<" "; cout<<endl
#define take(x,n)           for(int i=0;i<n;i++)  cin>>x[i];

#define watch(x) cout << (#x) << " is " << (x) << "\n"
#define watch2(x,y) cout <<(#x)<<" is "<<(x)<<" and "<<(#y)<<" is "<<(y)<<"\n"
#define watch3(x,y,z) cout <<(#x)<<" is "<<(x)<<" and "<<(#y)<<" is "<<(y)<<" and "<<(#z)<<" is "<<(z)<<"\n"

#define ll long long
#define ff first
#define ss second
#define null NULL
#define all(c) (c).begin(),(c).end()
#define nl "\n"

#define ld long double
#define eb emplace_back
#define pb push_back
#define pf push_front
#define mod 1000000007

typedef vector<ll> vl;
typedef vector< vl > vvl;
typedef pair< ll, ll> pll;
typedef map< ll, ll> mll;

ll fast_mod_exp(ll a, ll b, ll m) {
	ll res = 1;
	while (b > 0) {
		if (b & 1)
			res = (res * a) % m;

		a = (a * a) % m;
		b = b >> 1;
	}

	return res;
}

int main() {

	// Use ctrl+shift+b ( second option )
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

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

	ll n;
	cin >> n;

	ll m = 1000000007;

	ll v1 = fast_mod_exp(10, n, m);
	ll v2 = fast_mod_exp(9, n, m);
	ll v3 = fast_mod_exp(8, n, m);

	ll ans = (v1 + v3 - 2 * v2 + m) % m;

	cout << ans << nl;

	return 0;
}