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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i,N) for(int i=0;i<int(N);++i)
#define rep1(i,N) for(int i=1;i<int(N);++i)
#define all(a) (a).begin(),(a).end()
#define bit(k) (1LL<<(k))
#define SUM(v) accumulate(all(v), 0LL)

typedef pair<int, int> i_i;
typedef pair<ll, ll> l_l;
template <class T> using vec = vector<T>;
template <class T> using vvec = vector<vec<T>>;
struct fast_ios{ fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; }fast_ios_;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

#define TOSTRING(x) string(#x)
template <typename T> istream &operator>>(istream &is, vector<T> &vec) { for (T &x : vec) is >> x; return is; }
template <typename T> ostream &operator<<(ostream &os, const vector<T> &v) { os  << "["; for(auto _: v) os << _ << ", "; os << "]"; return os; };
template <typename T> ostream &operator<<(ostream &os, set<T> &st) { os << "("; for(auto _: st) { os << _ << ", "; } os << ")";return os;}
template <typename T, typename U> ostream &operator<<(ostream &os, const pair< T, U >& p){os << "{" <<p.first << ", " << p.second << "}";return os; }
template <typename T, typename U> ostream &operator<<(ostream &os, const map<T, U> &mp){ os << "["; for(auto _: mp){ os << _ << ", "; } os << "]" << endl; return os; }

#define DUMPOUT cerr
void dump_func(){ DUMPOUT << endl; }
template <class Head, class... Tail> void dump_func(Head &&head, Tail &&... tail) { DUMPOUT << head; if (sizeof...(Tail) > 0) { DUMPOUT << ", "; } dump_func(std::move(tail)...); }

#ifdef DEBUG
#define dbg(...) dump_func(__VA_ARGS__)
#define dump(...) DUMPOUT << string(#__VA_ARGS__) << ": "; dump_func(__VA_ARGS__)
#else
#define dbg(...)
#define dump(...)
#endif

const int INF = (ll)1e9;
const ll INFLL = (ll)1e18+1;
const ll MOD = 1000000007;
// const ll MOD = 998244353;
const long double PI = acos(-1.0);

/*
const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
const string dir = "DRUL";
*/
struct mint {
    long long x;
    mint(long long _x=0):x((_x%MOD+MOD)%MOD){}
    mint operator-() const { return mint(-x);}
    mint& operator+=(const mint a) {
        if ((x += a.x) >= MOD) x -= MOD;
        return *this;
    }
    mint& operator-=(const mint a) {
        if ((x += MOD-a.x) >= MOD) x -= MOD;
        return *this;
    }
    mint& operator*=(const mint a) {
        (x *= a.x) %= MOD;
        return *this;
    }
    mint operator+(const mint a) const {
        mint res(*this);
        return res+=a;
    }
    mint operator-(const mint a) const {
        mint res(*this);
        return res-=a;
    }
    mint operator*(const mint a) const {
        mint res(*this);
        return res*=a;
    }
    mint modpow(long long t) const {
        if (!t) return 1;
        mint a = modpow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }
    // for prime MOD
    mint inv() const {
        return modpow(MOD-2);
    }
    mint& operator/=(const mint a) {
        return (*this) *= a.inv();
    }
    mint operator/(const mint a) const {
        mint res(*this);
        return res/=a;
    }
    friend std::ostream& operator<<(std::ostream& os, const mint& a){
        os << a.x;
        return os;
    }
};
mint dp[1000100][2][2];
int main() {
    int N;
    cin >> N;
    rep(i,N+1)rep(j,2)rep(k,2)dp[i][j][k] = 0;
    dp[0][0][0] = 1;
    rep(i,N){
        rep(j,2)rep(k,2){
            for(int nxt = 0; nxt <= 9; nxt++){
                dp[i+1][j || nxt == 0][k || nxt == 9] += dp[i][j][k];
            }
        }
    }
    cout << dp[N][1][1] << endl;
}