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 <ctype.h>
#include <float.h>
#include <inttypes.h>
#include <limits.h>
#include <signal.h>
#include <stdarg.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <unistd.h>
#include <math.h>
#ifdef __cplusplus
#include <bits/stdc++.h>
#endif
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
//#pragma GCC optimize("unroll-loops")

//char rbuf[]; int rbufl;
//#define getchar() rbuf[rbufl++]
#define getchar getchar_unlocked
#define putchar putchar_unlocked
int32_t nextint(void){ char c=getchar(); while(c!='-'&&(c<'0'||'9'<c)) c=getchar(); bool s=false; if(c=='-'){s=true;c=getchar();} uint32_t x=0; while('0'<=c && c<='9'){ x=x*10+c-'0'; c=getchar(); } return s?-x:x; }
int64_t nextlong(void){ char c=getchar(); while(c!='-'&&(c<'0'||'9'<c)) c=getchar(); int s=0; if(c=='-'){s=1;c=getchar();} uint64_t x=0; while('0'<=c && c<='9'){ x=x*10+c-'0'; c=getchar(); } return s?-x:x; }
uint32_t nextstr(char *s){ char c=getchar(); while(c==' '||c=='\n') c=getchar(); uint32_t len=0; while(c!=' '&&c!='\n'){ *s++=c; len++; c=getchar(); } *s='\0'; return len; }
void printint(int32_t x){ if(x<0){x=-x; putchar('-');}else if(x==0){putchar('0');} int d=0; char s[10]; while(x){ s[d]='0'+x%10; x/=10; d++; } for (int i=d-1; i>=0; i--) putchar(s[i]); putchar('\n'); }
void rsorta_d32(void *base, const int count, const int size, const int index){
	typedef int32_t rsort_t;
	rsort_t (*a)[size] = (rsort_t(*)[size])base;
	rsort_t (*b)[size] = (rsort_t(*)[size])malloc(count*size*sizeof(rsort_t));
	int c[0x100];
	for(int k=0; k<4; k++){
		memset(c, 0, sizeof(c)); for(int i=0; i<count; i++){ c[(a[i][index]>>(k<<3))&0xFF]++; } for(int i=1; i<0x100; i++) c[i]+=c[i-1]; for(int i=count-1; i>=0; i--){ int to=--c[(a[i][index]>>(k<<3))&0xFF]; for(int j=0; j<size; j++){ b[to][j]=a[i][j]; } }{rsort_t(*t)[size]=a; a=b; b=t;}
	}
	free(b);
}

#define P 1000000007

int getk(int a, int x){
	if(x==1) return a;
	int b = getk(a, x/2);
	b = ((long)b)*b%P;
	if((x&1)==0){
		return b;
	}else{
		return ((long)b)*a%P;
	}
}

int main(void){
	int n=nextint();
	int ans = getk(10,n) + getk(8,n) - getk(9,n)*2;
	ans=(ans%P+P)%P;
	printf("%d\n", ans);
}