Time limit: 1.0 second	Memory limit: 64 MB

On the Day of the Flag of Russia a shop-owner decided to decorate the show-window of his shop with textile stripes of white, blue and red colors. He wants to satisfy the following conditions:

Stripes of the same color cannot be placed next to each other.

A blue stripe must always be placed between a white and a red or between a red and a white one.

Determine the number of the ways to fulfill his wish.

Example. For N = 3 result is following:

Problem illustration

Input

N, the number of the stripes, 1 ≤ N ≤ 45.

Output

M, the number of the ways to decorate the shop-window.

Sample

input

3

output

4

Problem Source:

2002-2003 ACM Central Region of Russia Quarterfinal Programming Contest, Rybinsk, October 2002

简单的递推，三种颜色的布，相邻的颜色不能相同，蓝色在红色与白色之间，则用1表示红色，2表示白色，Dp[i][j]表示第i个条纹的颜色为j是的状态数目，当j=1时，如果i-1的颜色为白色，则Dp[i][1]+=Dp[i-1][2],如果为蓝色，则取决于i-2的颜色为白色状态，所以Dp[i][1]=Dp[i-1][2]+Dp[i-2][2],则Dp[i][2]=Dp[i-1][1]+Dp[i-2][1].

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <cmath>

#include <string>

#include <queue>

#include <vector>

#include <deque>

#include <iostream>

#include <algorithm>

using namespace std;

long long Dp[50][3];

int main()

{

    memset(Dp,0,sizeof(Dp));

    Dp[1][1]=Dp[1][2]=1;

    for(int i=2;i<=45;i++)

    {

        Dp[i][1] =Dp[i-1][2]+Dp[i-2][2];

        Dp[i][2] = Dp[i-1][1]+Dp[i-2][1];

    }

    int n;

    while(~scanf("%d",&n))

    {

        cout<<Dp[n][1]+Dp[n][2]<<endl;

    }

    return 0;

}