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 <iostream>
#include <iomanip>
#include <vector>
#include <set>
#include <string>
#include <queue>
#include <algorithm>
#include <map>
#include <cmath>
#include <numeric>
#include <list>
#include <stack>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <tuple>
#include <deque>
#include <complex>
#include <bitset>

using namespace std;

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> pii;
typedef vector<pii> vpii;

typedef long long ll;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef pair<long long, long long> pll;
typedef vector<pll> vpll;

typedef long double ld;
typedef vector<ld> vld;

typedef vector<bool> vb;

#define rep(i, n) for (ll i = 0; i < (n); i++)
#define reps(i, n) for (ll i = 1; i <= (n); i++)
#define rrep(i, n) for (ll i = (n) - 1; i >= 0; i--)
#define rreps(i, n) for (ll i = (n); i >= 1; i--)
#define all(v) (v).begin(), (v).end()

template <class T> void chmin(T& a, T b) { a = min(a, b);}
template <class T> void chmax(T& a, T b) { a = max(a, b);}

constexpr int INF = 1 << 30;
constexpr ll INFL = 1LL << 60;
constexpr ll MOD = 1000000007;
constexpr ld EPS = 1e-12;
ld PI = acos(-1.0);

struct mint {
    static const long long mod = 1000000007;
    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;}

    friend const mint operator+(const mint& a, const mint& b) { mint ret = a; return ret += b; }
    friend const mint operator-(const mint& a, const mint& b) { mint ret = a; return ret -= b; }
    friend const mint operator*(const mint& a, const mint& b) { mint ret = a; return ret *= b; }
    friend ostream& operator<<(ostream& s, const mint& a) { return s << a.x; }

    mint pow(long long t) const {
        if (!t) return 1;
        mint a = pow(t >> 1);
        a *= a;
        if (t & 1) a *= *this;
        return a;
    }

    // for prime mod
    mint inv() const { return pow(mod - 2);}
    mint& operator/=(const mint a) { return (*this) *= a.inv();}
    friend const mint operator/(const mint& a, const mint& b) { mint ret = a; return ret /= b; }
};

void solve() {
    ll n;
    cin >> n;
    mint ans = mint(10).pow(n) - 2 * mint(9).pow(n) + mint(8).pow(n);
    cout << ans << endl;
    return;
}

int main() {
    std::cin.tie(nullptr);
    std::ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    solve();
}