Kolya and Tanya

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 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 109 + 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):

思路：

The number of ways to choose ai (without any conditions) — 33n. Let A be the number of ways to choose ai with condition that in every triangle gnomes have 6 coins. Then answer is 33n - A.

Note that we can separate all gnomes on n independent triangles (i-th triangle contains of ai, ai + n, ai + 2n, i < n). So we can count answer for one triangle and then multiply the results. To count answer for one triangle, we can note that there are 7 ways to choose gnomes with 6 coins (all permutations 1, 2, 3 and 2, 2, 2).

So answer is — 33n - 7n. We count it in O(n).

   组合数学题，一开始Wa了，原来是在乘一个数时发生了溢出，所以要用long long。

AC代码：

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