【Codeforces Round 324 (Div 2)B】【容斥】Kolya and Tanya 环上n个3元组至少有一组和不为6

B. Kolya and Tanya

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Kolya loves putting gnomes at the circle table and giving them coins, and Tanya loves studying triplets of gnomes, sitting in the vertexes of an equilateral triangle.

More formally, there are 3n gnomes sitting in a circle. Each gnome can have from 1 to 3 coins. Let's number the places in the order they occur in the circle by numbers from 0 to 3n - 1, let the gnome sitting on the i-th place have a_i coins. If there is an integer i (0 ≤ i < n) such that a_i + a_i + n + a_i + 2n ≠ 6, then Tanya is satisfied.

Count the number of ways to choose a_i so that Tanya is satisfied. As there can be many ways of distributing coins, print the remainder of this number modulo 10⁹ + 7. Two ways, a and b, are considered distinct if there is index i (0 ≤ i < 3n), such that a_i ≠ b_i (that is, some gnome got different number of coins in these two ways).

Input

A single line contains number n (1 ≤ n ≤ 10⁵) — the number of the gnomes divided by three.

Output

Print a single number — the remainder of the number of variants of distributing coins that satisfy Tanya modulo 10⁹ + 7.

Sample test(s)

input

1

output

20

input

2

output

680

Note

20 ways for n = 1 (gnome with index 0 sits on the top of the triangle, gnome 1 on the right vertex, gnome 2 on the left vertex):

#include<stdio.h>
#include<string.h>
#include<ctype.h>
#include<math.h>
#include<iostream>
#include<string>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre(){freopen("c://test//input.in","r",stdin);freopen("c://test//output.out","w",stdout);}
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1,class T2>inline void gmax(T1 &a,T2 b){if(b>a)a=b;}
template <class T1,class T2>inline void gmin(T1 &a,T2 b){if(b<a)a=b;}
const int N=0,M=0,Z=1e9+7,ms63=1061109567;
int casenum,casei;
LL mul(LL x,int p)
{
	LL y=1;
	while(p)
	{
		if(p&1)y=y*x%Z;
		x=x*x%Z;
		p>>=1;
	}
	return y;
}
int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		printf("%lld\n",(mul(27,n)+Z-mul	(7,n))%Z);
	}
	return 0;
}
/*
【题意】
有3n（1<=n<=1e5）个点，每个点的数可以取（1，2，3）编号从0到3n-1.构成一个环。
问你在所有可能的环中有多少个环，使得至少存在一个p（0<=p<n），满足a[p]+a[n+p]+a[n+n+p]!=6

【类型】
简单容斥

【分析】
本来环的总是一共是3^3n。
对于单一的三元组，认定其非法的条件是a[p]+a[n+p]+a[n+n+p]=6，即（123）全排列6种+（222）一种
于是非法的环，每个三个都要是非法的，于是数量便是7^n。
这样答案就算出来了，是27^n-7^n

*/