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 Rust (1.42.0) (judge verdict: Accepted)

//use itertools::Itertools;
use std::cmp;
use std::collections::BTreeMap;
use std::collections::BTreeSet;
use std::collections::BinaryHeap;
use std::collections::HashMap;
use std::collections::HashSet;
use std::collections::VecDeque;
use std::io::Read;
use std::usize::MAX;

macro_rules! input {(source = $s:expr, $($r:tt)*) => {let mut iter = $s.split_whitespace();let mut next = || { iter.next().unwrap() };input_inner!{next, $($r)*}};($($r:tt)*) => {let stdin = std::io::stdin();let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));let mut next = move || -> String{bytes.by_ref().map(|r|r.unwrap() as char).skip_while(|c|c.is_whitespace()).take_while(|c|!c.is_whitespace()).collect()};input_inner!{next, $($r)*}};}
macro_rules! input_inner {($next:expr) => {};($next:expr, ) => {};($next:expr, $var:ident : $t:tt $($r:tt)*) => {let $var = read_value!($next, $t);input_inner!{$next $($r)*}};}
macro_rules! read_value {($next:expr, ( $($t:tt),* )) => {( $(read_value!($next, $t)),* )};($next:expr, [ $t:tt ; $len:expr ]) => {(0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()};($next:expr, chars) => {read_value!($next, String).chars().collect::<Vec<char>>()};($next:expr, usize1) => {read_value!($next, usize) - 1};($next:expr, [ $t:tt ]) => {{let len = read_value!($next, usize);(0..len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()}};($next:expr, $t:ty) => {$next().parse::<$t>().expect("Parse error")};}

fn solve() {
    input! {
        n: usize,
    }

    if n < 2 {
        println!("0");
        return;
    }

    let mut ans = 0;
    ans = power_mod(10, n as u64, LARGE_PRIME);
    ans += LARGE_PRIME;
    // 0も9もなし
    ans += power_mod(8, n as u64, LARGE_PRIME);
    ans += LARGE_PRIME;
    // 0のみあり
    ans -= power_mod(9, n as u64, LARGE_PRIME);
    ans += LARGE_PRIME;
    // 9のみあり
    ans -= power_mod(9, n as u64, LARGE_PRIME);
    ans += LARGE_PRIME;

    ans %= LARGE_PRIME;

    println!("{}", ans);
}

fn main() {
    let thd = std::thread::Builder::new().stack_size(104_857_600);
    thd.spawn(|| solve()).unwrap().join().unwrap();

    /*
    // 入力を一括で読み込む場合
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();
    let mut input = buf.split_whitespace();

    // inputに対しnext()で読み込んでいく
    let q: usize = input.next().unwrap().parse().unwrap();
    */
}

const LARGE_PRIME: u64 = 1_000_000_007;

// @〜でsnippet
fn power_mod(base: u64, exp: u64, modulo: u64) -> u64 {
    if exp == 0 {
        return 1;
    } else if exp & 1 == 0 {
        return (power_mod(base, exp / 2, modulo) % modulo).pow(2) % modulo;
    } else {
        return power_mod(base, exp - 1, modulo) % modulo * base % modulo;
    }
}

fn inverse_mod(element: u64, modulo: u64) -> u64 {
    // フェルマーの小定理からpが素数なら a^(p-1) = 1
    // よってa*a^(p-2) = 1 より a^(p-2)がaの逆元
    power_mod(element, modulo - 2, modulo) % modulo
}

fn permutation_mod(m: u64, n: u64, modulo: u64) -> u64 {
    // m P n = m! / (m - n)!
    //       = m*(m-1)*(m-2)*...*(m-n+1)
    let mut numerator = 1;
    for i in 0..n {
        numerator *= m - i;
        numerator %= modulo;
    }

    numerator
}
fn combination_mod(m: u64, n: u64, modulo: u64) -> u64 {
    // m C n = m! / ( n! * (m - n)! )
    //       = m*(m-1)*(m-2)*...*(m-n+1)/n!
    //       = m*(m-1)*(m-2)*...*(m-n+1) * (n!)^-1
    let permutation = permutation_mod(m, n, modulo);

    // n!
    let mut element = 1;
    for j in 1..(n + 1) {
        element *= j;
        element %= modulo;
    }
    permutation * inverse_mod(element, modulo) % modulo
}