codeforces166E

Tetrahedron

 CodeForces - 166E 

You are given a tetrahedron. Let's mark its vertices with letters A, B, C and Dcorrespondingly.

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).

Examples

Input

2

Output

3

Input

4

Output

21

Note

The required paths in the first sample are:

• D - A - D
• D - B - D
• D - C - D

sol：直接dp丝毫不慌，dp[i][0/1/2/3]表示第i步，当前位于节点j的方案数

#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=0;
    bool f=0;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0)
    {
        putchar('-'); x=-x;
    }
    if(x<10)
    {
        putchar(x+'0'); return;
    }
    write(x/10);
    putchar((x%10)+'0');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('
')
const int N=100005;
int n;
int a[N],A[N],Len[N];
int main()
{
    int i,j,ans=1;
    R(n);
    if(n<=2) {Wl(n); return 0;}
    for(i=1;i<=n;i++)
    {
        R(a[i]);
    }
    *A=0;
    for(i=1;i<=n;i++)
    {
        if(a[i]>0)
        {
            A[++*A]=a[i];
            Len[*A]=1;
        }
        else
        {
            A[++*A]=0;
            for(;i<=n&&a[i]==0;i++) Len[*A]++;
            i--;
        }
    }
    for(i=1;i<=n;i++) ans=max(ans,Len[i]);
    if(*A==1) ans=n;
    if(*A==2)
    {
        if(A[1]==0) ans=max(Len[1],Len[2]+1);
        else ans=Len[2];
    }
    for(i=1;i<=(*A)-2;i++)
    {
        int tmp;
        if(A[i]==0)
        {
            tmp=Len[i+1]+1;
        }
        else if(A[i+1]==0)
        {
            if(Len[i+1]==1) tmp=Len[i+1]+1;
            else
            {
                tmp=1+Len[i+2];
                for(j=i+3;j<=*A;j++)
                {
                    if(A[j]==A[j-1]+A[j-2]) tmp+=Len[j];
                    else break;
                }
                ans=max(ans,tmp);
                continue;
            }
        }
        else tmp=Len[i]+Len[i+1];
        for(j=i+2;j<=*A;j++)
        {
            if(A[j]==A[j-1]+A[j-2]) tmp+=Len[j];
            else break;
        }
        ans=max(ans,tmp);
    }
    Wl(ans);
    return 0;
}
/*
input
10
1 2 3 5 8 13 21 34 55 89
output
10

input
5
1 1 1 1 1
output
2

input
10
1 1 0 0 0 0 0 0 0 1
output
7
*/