Codeforces Round 324 (Div. 2)
B
Kolya and Tanya
Codeforces
1500
原题
← AC →
翻译中…

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.

翻译中…

More formally, there are $3*n*$ 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 $3*n* - 1$, let the gnome sitting on the $*i*$-th place have $*a*_{*i*}$ coins. If there is an integer $*i*$ ($0 ≤ *i* < *n*$) such that $*a*_{*i*} + *a*_{*i* + *n*} + *a*_{*i* + 2*n*} ≠ 6$, then Tanya is satisfied.

翻译中…

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 $10^{9} + 7$. Two ways, $*a*$ and $*b*$, are considered distinct if there is index $*i*$ ($0 ≤ *i* < 3*n*$), such that $*a*_{*i*} ≠ *b*_{*i*}$ (that is, some gnome got different number of coins in these two ways).

翻译中…

Input

翻译中…

A single line contains number $*n*$ ($1 ≤ *n* ≤ 10^{5}$) — the number of the gnomes divided by three.

翻译中…

Output

翻译中…

Print a single number — the remainder of the number of variants of distributing coins that satisfy Tanya modulo $10^{9} + 7$.

翻译中…

Note

翻译中…

$20$ 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): image

翻译中…

样例

Input #1
1
Output #1
20
InputCopy OutputCopy
Input #2
2
Output #2
680
InputCopy OutputCopy
题目信息
题号
584B
时间限制
1 s
内存限制
256 MB
标签
组合数学

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