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)

#include <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <vector>
#define m_p make_pair
#define sz(x) (int)x.size()
#define line cerr<<"--------------------\n";
#define see(x) cerr<<x<<" "
#define seeln(x) cerr<<x<<endl
#define out(x) cerr<<#x<<" = "<<x<<" "
#define outln(x) cerr<<#x<<" = "<<x<<endl
#define outarr(x,l,r) cerr<<#x"["<<l<<"-"<<r<<"] = "; for (int _i=l;_i<=r;++_i) cerr<<x[_i]<<" ";cerr<<endl;
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
#define gc() getchar()
//char buf[1<<23],*p1=buf,*p2=buf;
//#define gc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
template <class T> void read(T &x)
{
    x=0; char c=gc(); int flag=0;
    while (c<'0'||c>'9') flag|=(c=='-'),c=gc();
    while (c>='0'&&c<='9') x=(x<<3)+(x<<1)+(c^48),c=gc();
    if (flag) x=-x;
}
template <class T> inline T _max(T a,T b){return a>b ? a : b;}
template <class T> inline T _min(T a,T b){return a<b ? a : b;}
template <class T> inline bool checkmax(T &a,T b){return b>a ? a=b,1 : 0;}
template <class T> inline bool checkmin(T &a,T b){return b<a ? a=b,1 : 0;}
const int N=10005,mod=1000000000+7;
int Add(int a,int b){a+=b; return a>=mod ? a-mod : a;}
int Sub(int a,int b){a-=b; return a>=0 ? a : a+mod;}
int Mul(int a,int b){return 1LL*a*b%mod;}
void add(int &a,int b){a=Add(a,b);}
void sub(int &a,int b){a=Sub(a,b);}
void mul(int &a,int b){a=Mul(a,b);}
int qpow(int a,int b)
{
	int ans=1;
	while (b)
	{
		if (b&1) mul(ans,a);
		b>>=1;
		mul(a,a);
	}
	return ans;
}
int n;
void init()
{
	read(n);
}

void solve()
{
	if (n==1)
	{
		printf("0\n");
		return;
	}
	int ans=qpow(10,n);
	add(ans,qpow(8,n));
	sub(ans,qpow(9,n));
	sub(ans,qpow(9,n));
	printf("%d\n",ans);
}

int main()
{
	init();
	solve();
	return 0;
}