Codeforces Round #324 (Div. 2) B

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.

Examples

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):

题意： 输入一个n      3*n个位置 0~3n-1 每个位置的k等于（1，2，3）

若a_i + a_i + n + a_i + 2n ≠ 6 则算做一种情况  问共有多少情况 % 1000000000+9

题解：   计算式子

ans1=3^3n%mod

ans2=7^n%mod

( ans1-ans2)%mod

若 ans1<ans2

则输出（ans1+mod-ans2）%mod

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<queue>
 #include<stack>
 #include<cmath>
 #define ll __int64
 #define pi acos(-1.0)
 #define mod 1000000007
 using namespace std;
 ll n;
 ll ans1,ans2;
 ll quickmod(ll a,ll b)
 {
     ll sum=;
     while(b)
     {
         if(b&)
             sum=(sum*a)%mod;
         b>>=;
         a=(a*a)%mod;
     }
     return sum;
 }
 int main()
 {
     scanf("%I64d",&n);
     ans1=quickmod(,*n)%mod;
     ans2=quickmod(,n)%mod;
     if(ans1>=ans2)
     printf("%I64d\n",(ans1-ans2)%mod);
     else
     printf("%I64d\n",(ans1+mod-ans2)%mod);
     return ;
 }