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 PHP (7.4.4) (judge verdict: Accepted)

<?php
const MOD = 10 ** 9 + 7;
list($n) = ints();
echo $n >= 2 ? modSub(modMul(2, modSub(modPow(10, $n), modPow(9, $n))), modSub(modPow(10, $n), modPow(8, $n))) : 0;
function ints()
{
    return array_map('intval', explode(' ', trim(fgets(STDIN))));
}
// 足し算
function modAdd($x, $y)
{
    return ($x + $y) % MOD;
}

// SUM
function modSum(...$xs)
{
    $res = 0;
    foreach ($xs as $x) $res = modAdd($res, $x);
    return $res;
}

// 引き算
function modSub($x, $y)
{
    return ($x + MOD - $y) % MOD;
}

// 掛け算
function modMul($x, $y)
{
    return ($x * $y) % MOD;
}

// PRODUCT
function modProd(...$xs)
{
    $res = 1;
    foreach ($xs as $x) $res = modMul($res, $x);
    return $res;
}

// 割り算
function modDiv($x, $y)
{
    return modMul($x, modPow($y, MOD - 2));
}

// 累乗（繰り返し二乗法）
function modPow($n, $x)
{
    if ($x === 0) return 1;
    $res = (modPow($n, $x >> 1) ** 2) % MOD;
    if ($x % 2 === 1) $res = modMul($res, $n);
    return $res;
}

// 階乗
function modFac($n)
{
    if ($n === 0) return 1;
    return modMul($n, modFac($n - 1));
}

// 順列
function nPr($n, $r)
{
    if ($r === 0) return 1;
    return modMul(nPr($n, $r - 1), $n - $r + 1);
}

// 組み合わせ
function nCr($n, $r)
{
    $r = min($r, $n - $r);
    if ($r === 0) return 1;
    return modDiv(nPr($n, $r), modFac($r));
}

// 階乗、逆元を前処理
function init($n)
{
    global $fact, $ifact;
    $fact[0] = 1;
    for ($i = 1; $i <= $n; ++$i) $fact[$i] = modMul($fact[$i - 1], $i);
    $ifact[$n] = modDiv(1, $fact[$n]);
    for ($i = $n; $i >= 1; --$i)
        $ifact[$i - 1] = modMul($ifact[$i], $i);
}

// 順列（前処理）
function nPr2($n, $r)
{
    global $fact, $ifact;
    return modMul($fact[$n], $ifact[$n - $r]);
}

// 組み合わせ（前処理）
function nCr2($n, $r)
{
    global $fact, $ifact;
    return modMul(modMul($fact[$n], $ifact[$r]), $ifact[$n - $r]);
}