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>
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
using namespace std;
// using namespace __gnu_pbds;
#define lli long long int
#define llu unsigned long long int
#define pb push_back
#define rt return 0
#define endln "\n"
#define all(x) x.begin(), x.end()
#define sz(x) (lli)(x.size())
const lli MOD = 1e9 + 7;
const double PI = 2 * acos(0.0);
// cout << fixed << setprecision(0) << pi <<endl;
// typedef tree<int, null_type, less<int>, rb_tree_tag,
//              tree_order_statistics_node_update>
//     new_data_set;
// for multiset
// typedef tree<int, null_type, less_equal<int>, rb_tree_tag,
//              tree_order_statistics_node_update>
//     new_data_set;
// order_of_key(val): returns the number of values less than val
// find_by_order(k): returns an iterator to the kth largest element (0-based)

lli power(lli x, lli y) {
    // log y complexity
    // return (x^y) % MOD
    lli ans = 1;
    x = x % MOD;
    while (y > 0) {
        if (y & 1) {
            ans *= x;
            ans %= MOD;
        }
        y = y >> 1;
        x = (x * x) % MOD;
    }
    return ans;
}

void solve() {
    lli n, a;
    cin >> n;
    lli ans = (power(9, n) * 2) % MOD;
    ans -= power(8, n);
    ans += MOD;
    ans %= MOD;
    lli tot = power(10, n);

    tot -= ans;
    tot += MOD;
    tot %= MOD;
    cout << tot << endl;
}

int main(void) {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    // lli t; cin >> t; while(t--)
    solve();
    rt;
}