codeforces 584B Kolya and Tanya

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 ai coins. If there is an integer i (0 ≤ i < n) such that ai + ai + n + ai + 2n ≠ 6, then Tanya is satisfied.

Count the number of ways to choose ai so that Tanya is satisfied. As there can be many ways of distributing coins, print the remainder of this number modulo 1e9 + 7. Two ways, a and b, are considered distinct if there is index i (0 ≤ i < 3n), such that ai ≠ bi (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):

题目大意：给出n个等边三角形，每个顶点都可以是1~3中的一个数，一个等边三角形三个顶点的总和不能是6。在n个三角形中只要有一个等边三角形满足条件，当前情况就是合格的，问有多少种合格的情况。

解题思路：所有的情况27n，没有一个三角形满足条件7n，答案27n−7n。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
using namespace std;

typedef long long ll;
const ll mod = 1e9 + 7;

ll quick_mod(ll a, ll b) {
ll ans = 1;
while(b) {
if(b&1) ans = (ans * a) % mod;
b >>= 1;
a = (a * a) % mod;
}
return ans;
}

int main() {
ll n;
scanf("%lld", &n);
ll A = quick_mod(27, n);
ll B = quick_mod(7, n);
printf("%lld\n", ((A % mod - B % mod) + mod) % mod);
return 0;
}