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)

// abc178_c
#include <bits/stdc++.h>
#ifdef LOCAL
#include "../cxx-prettyprint/prettyprint.hpp"
#endif
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;

#define REP(i, n) for (int (i) = 0 ; (i) < (int)(n) ; ++(i))
#define REPN(i, m, n) for (int (i) = m ; (i) < (int)(n) ; ++(i))
#define REP_REV(i, n) for (int (i) = (int)(n) - 1 ; (i) >= 0 ; --(i))
#define REPN_REV(i, m, n) for (int (i) = (int)(n) - 1 ; (i) >= m ; --(i))
#define ALL(x) x.begin(), x.end()

#define INF (ll)(1e9)
#define MOD (1000000007)

#define print2D(h, w, arr) REP(i, h) { REP(j, w) cout << arr[i][j] << " "; cout << endl; }
#define print_line(vec, n) {for(int idx=0;idx<(n-1);idx++) cout << (vec)[idx] << " "; cout << (vec)[(n)-1] << endl;}
template<class T> void print(const T& x){cout << x << "\n";}
template<class T, class... A> void print(const T& first, const A&... rest) { cout << first << " "; print(rest...); }
struct PreMain {PreMain(){cin.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(20);}} premain;


const int mod = 1e9 + 7;
struct mint {
    ll x;
    mint(ll a=0){x = a>=0 ? a%mod : mod-(-a)%mod;}
    mint(const mint& m):x(m.x){}
    mint& operator+=(const mint& a) { if ((x += a.x) >= mod) x -= mod; return *this; }
    mint& operator+=(const int& a) { if ((x += a) >= mod) x -= mod; return *this; }
    mint& operator-=(const mint& a) { if ((x += mod-a.x) >= mod) x -= mod; return *this; }
    mint& operator-=(const int& a) { if ((x += mod-a) >= mod) x -= mod; return *this; }
    mint& operator*=(const mint& a) { (x *= a.x) %= mod; return *this; }
    mint& operator*=(const int& a) { (x *= a) %= mod; return *this; }
    mint operator+(const mint& a) const { mint res(*this); return res+=a; }
    mint operator+(const int& a) const { mint res(*this); return res+=a; }
    mint operator-(const mint& a) const { mint res(*this); return res-=a; }
    mint operator-(const int& a) const { mint res(*this); return res-=a; }
    mint operator-() const {mint res(*this); return mint(0)-res;}
    mint operator*(const mint& a) const { mint res(*this); return res*=a; }
    mint operator*(const int& a) const { mint res(*this); return res*=a; }
    bool operator==(const mint& a) const { return x == a.x;}
    bool operator==(const int& a) const { return x == a;}
    bool operator!=(const mint& a) const { return x != a.x;}
    bool operator!=(const int& a) const { return x != a;}
    operator bool() const {return x>0;}

    mint pow(ll t) const {mint r=1,a=*this; do{if(t&1)r*=a;a*=a;}while(t>>=1);return r;}
    mint pow(mint t) const {mint r=1,a=*this; do{if(t.x&1)r*=a;a*=a;}while(t.x>>=1);return r;}
    mint inv() const { return pow(mod-2); }
    mint& operator/=(const mint& a) { return (*this) *= a.inv(); }
    mint& operator/=(const int& a) { return (*this) *= mint(a).inv(); }
    mint operator/(const mint& a) const { mint res(*this); return res/=a; }
    mint operator/(const int& a) const { mint res(*this); return res/=a; }

    friend ostream& operator<<(ostream& os, const mint& m) {cout << m.x; return os;}
    friend istream& operator>>(istream& is, mint& m) { return is >> m.x; }
};

int main() {
#ifdef LOCAL
    ifstream in("../arg.txt"); cin.rdbuf(in.rdbuf());
#endif

    int N;
    cin >> N;

    mint x = mint(10).pow(N);
    mint y = mint(9).pow(N);
    mint z = mint(9).pow(N);
    mint w = mint(8).pow(N);

    mint ans = x - y - z + w;
    print(ans);


    return 0;
}