Kolya and Tanya (次幂取模)

Kolya and Tanya

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Description

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 Input

Input

1

Output

20

Input

2

Output

680

Hint

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 <cstdio>
typedef long long int LL;
const LL MOD = 1000000000+7;
LL deal_pow(LL a, LL b)
{
    LL ans=1;
    while(b)
    {
        if(b& 1) 
            ans=(ans*a)%MOD;
        b/= 2;
        a=(a*a)%MOD;
    }
    return ans;
}
int main()
{
    LL n; 
    while(scanf("%lld", &n) != EOF)
    {
        printf("%I64d
", (deal_pow(3, 3*n)-deal_pow(7, n)+MOD)%MOD);   //次幂取模后相减会出现负数 ; 
    }
    return 0;
}