You are given a tetrahedron.

最新推荐文章于 2024-10-16 20:49:11 发布

fisty

最新推荐文章于 2024-10-16 20:49:11 发布

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分类专栏： ------基础DP 文章标签： ACM

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本文探讨了一个在四面体中蚂蚁从顶点出发，在边间移动恰好 n 步后返回初始顶点的不同路径数量计算问题，并提供了解决方法及代码实现。

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You are given a tetrahedron.

Let's mark its vertices with letters A, B, C and D correspondingly.

$\text{[math]}$

An ant is standing in the vertex D of the tetrahedron. The ant is quite active and he wouldn't stay idle. At each moment of time he makes a step from one vertex to another one along some edge of the tetrahedron. The ant just can't stand on one place.

You do not have to do much to solve the problem: your task is to count the number of ways in which the ant can go from the initial vertex D to itself in exactly n steps. In other words, you are asked to find out the number of different cyclic paths with the length of n from vertex D to itself. As the number can be quite large, you should print it modulo 1000000007 (10⁹ + 7).

Input

The first line contains the only integer n (1 ≤ n ≤ 10⁷) — the required length of the cyclic path.

Output

Print the only integer — the required number of ways modulo 1000000007 (10⁹ + 7).

Sample Input

Input

Output

Input

Output

Hint

The required paths in the first sample are:

D - A - D
D - B - D
D - C -D

/***********************************************
 * Author: fisty
 * Created Time: 2015/2/8 19:24:29
 * File Name   : 6_B.cpp
 *********************************************** */
#include <iostream>
#include <cstring>
#include <deque>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <string>
#include <vector>
#include <cstdio>
#include <bitset>
#include <algorithm>
using namespace std;
#define MOD  1000000007
int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n;
    long long  f[10000010] = {0, 0, 3, 6};
    cin >> n;
    for(int i = 4;i <= n; i++){
        f[i] = (3 * f[i-2] + 2 * f[i-1]) % MOD;
    }
    cout << f[n] << endl;
    return 0;
}