题意：给出一个$n\times m$的方格图，现在要用如下L型的占3个的积木拼到这个图中,总共有多少种拼法使图满。

#include<bits/stdc++.h>
using namespace std;
long long n,m,k=1,Now;
int Mod=1000000007;
struct Matrix
{long long a[129][129];Matrix(){for(long long i=0;i<k;i++)a[i][i]=1;}Matrix(int x){memset(a,0,sizeof(a));}Matrix operator*(Matrix& B)//矩阵乘法 {Matrix t(0);for(int i=0;i<k;i++)for(int j=0;j<k;j++)for(int l=0;l<k;l++)t.a[i][j]=(t.a[i][j]+a[i][l]*B.a[l][j])%Mod;return t;}
}; 	
Matrix quickmul(Matrix ans,long long p)//矩阵快速幂 
{Matrix t;	while(p){if(p&1)t=t*ans;ans=ans*ans;p>>=1;}return t;
}
Matrix A(0);
void dfs(int now,int next)
{//cout<<now<<" "<<next<<endl;if(now+1==k){A.a[Now][next]++;return ;}for(int l=0;l<m;l++){if((now>>l)&1)continue;//当前列的位置上有拼图则跳过int temp=1<<l;int now1,next1;//递归,四种拼图 now1=temp|(temp<<1);// 1 1next1=temp;         // 0 1if((l+1<m)&&!(now&now1||next&next1))dfs(now1|now,next1|next);now1=temp|(temp<<1);// 1 1 next1=temp<<1;      // 1 0if((l+1<m)&&!(now&now1||next&next1))dfs(now1|now,next1|next);now1=temp;		    // 0 1 next1=temp|temp<<1; // 1 1if((l+1<m)&&!(now&now1||next&next1))dfs(now1|now,next1|next);now1=temp;	        // 1 0 next1=temp|temp>>1; // 1 1if((l)&&!(now&now1||next&next1))dfs(now1|now,next1|next);return ;}
}
int main()
{cin>>n>>m;for(long long i=0;i<m;i++)k<<=1;for(Now=0;Now<k;Now++)dfs(Now,0);Matrix ANS=quickmul(A,n);	/*for(int i=0;i<k;i++){for(int j=0;j<k;j++)cout<<ANS.a[i][j]<<" ";cout<<endl;	}*/cout<<ANS.a[0][0]<<endl;return 0;
}