Poor God Water （矩阵快速幂）

God Water likes to eat meat, fish and chocolate very much, but unfortunately, the doctor tells him that some sequence of eating will make them poisonous.

Every hour, God Water will eat one kind of food among meat, fish and chocolate. If there are 3 continuous hours when he eats only one kind of food, he will be unhappy. Besides, if there are 3 continuous hours when he eats all kinds of those, with chocolate at the middle hour, it will be dangerous. Moreover, if there are 3 continuous hours when he eats meat or fish at the middle hour, with chocolate at other two hours, it will also be dangerous.

Now, you are the doctor. Can you find out how many different kinds of diet that can make God Water happy and safe during NNN hours? Two kinds of diet are considered the same if they share the same kind of food at the same hour. The answer may be very large, so you only need to give out the answer module 1000000007

Input

The fist line puts an integer T that shows the number of test cases. (T≤1000)

Each of the next T lines contains an integer N that shows the number of hours. (1≤N≤)

Output

For each test case, output a single line containing the answer.

样例输入

3
3
4
15

样例输出

20
46
435170

题意：有N个小时，有三种食物（用1 ，2 ，3代替好了），每个小时要吃一种食物，要求任意连续三个小时不能出现111，222，333，132，231，313，323的方案数

题解：dp[i][j]表示最后两种食物分别为i，j的方案数，转移就是在增加一种食物，可以根据要求推出下面的递推矩阵.

然后直接套用矩阵快速幂的模板即可

#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define linf 0x3f3f3f3f3f3f3f3fLL
#define pi acos(-1.0)
#define ms(a,b) memset(a,b,sizeof(a))
#define max3(a, b, c) max(a, max(b, c))
#define maxn 1e5+7
using namespace std;
typedef long long ll;
ll qpow(ll x, ll y, ll mod){ll s=1;while(y){if(y&1)s=s*x%mod;x=x*x%mod;y>>=1;}return s;}
inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();return x*f;}
ll qpow(ll a, ll b){ll s=1;while(b>0){if(b%2==1)s=s*a;a=a*a;b=b>>1;}return s;}

const int N=9;
struct Matrix{
    ll matrix[N][N];
};

const int mod = 1e9 + 7;

void init(Matrix &res)
{
    memset(res.matrix,0,sizeof(res.matrix));
    for(int i=0;i<N;i++)
        res.matrix[i][i]=1;
}
Matrix multiplicative(Matrix a,Matrix b)
{
    Matrix res;
    memset(res.matrix,0,sizeof(res.matrix));
    for(int i = 0 ; i < N ; i++){
        for(int j = 0 ; j < N ; j++){
            for(int k = 0 ; k < N ; k++){
                res.matrix[i][j] += a.matrix[i][k]*b.matrix[k][j];
                res.matrix[i][j] %= mod;
            }
        }
    }
    return res;
}
Matrix Pow(Matrix mx,ll m)
{
    Matrix res,base=mx;
    init(res);
    while(m)
    {
        if(m&1)
            res=multiplicative(res,base);
        base=multiplicative(base,base);
        m>>=1;
    }
    return res;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        ll n,ant = 0;
        scanf("%lld",&n);
        if(n == 1)  printf("3\n");
        else if(n == 2) printf("9\n");
        else
        {
            Matrix res,ans = {
                0,0,0, 1,0,0, 1,0,0,
                1,0,0, 0,0,0, 1,0,0,
                1,0,0, 1,0,0, 1,0,0,

                0,1,0, 0,1,0, 0,0,0,
                0,1,0, 0,0,0, 0,1,0,
                0,0,0, 0,1,0, 0,1,0,

                0,0,1, 0,0,1, 0,0,1,
                0,0,1, 0,0,0, 0,0,1,
                0,0,1, 0,0,1, 0,0,0
            };
            res=Pow(ans,n-2);

            for(int i = 0;i < N;i++)
                for(int j = 0;j < N;j++)
                    ant = (ant + res.matrix[i][j]) % mod;
            printf("%lld\n",ant);
        }
    }
    return 0;
}