#include<bits/stdc++.h>
#define ll long long
#define fo(i,x,y) for(int i=x;i<=y;i++)
#define fd(i,x,y) for(int i=x;i>=y;i--)
using namespace std;

const int maxn=3007,mo=1e9+7;

int n,m,Ans;
int f[maxn][maxn][2];

void Init(){
	scanf("%d%d",&n,&m);
}

void Plus(int &a,int b){
	(a+=b)%=mo;
}
void Solve(){
	f[0][0][1]=1;
	fo(i,1,n) f[0][i][0]=1;
	fo(i,0,m-1){
		fo(j,0,n)
			fo(k,0,1){
				if (j>0){
					Plus(f[i+1][j-1][k|(j-1==0)],f[i][j][k]);
					Plus(f[i+1][j][k|(j-1==0)],f[i][j][k]);
				}
				if (j<n){
					Plus(f[i+1][j+1][k],f[i][j][k]);
					Plus(f[i+1][j][k],f[i][j][k]);
				}
			}
	}
}

void Print(){
	fo(i,0,n) Plus(Ans,f[m][i][1]);
	printf("%d\n",Ans);
}

int main(){
	Init();
	Solve();
	Print();
	return 0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

Joisino has a lot of red and blue bricks and a large box. She will build a tower of these bricks in the following manner.

First, she will pick a total of N bricks and put them into the box. Here, there may be any number of bricks of each color in the box, as long as there are N bricks in total. Particularly, there may be zero red bricks or zero blue bricks. Then, she will repeat an operation M times, which consists of the following three steps:

* Take out an arbitrary brick from the box.
* Put one red brick and one blue brick into the box.
* Take out another arbitrary brick from the box.



After the M operations, Joisino will build a tower by stacking the 2 \times M bricks removed from the box, in the order they are taken out. She is interested in the following question: how many different sequences of colors of these 2 \times M bricks are possible? Write a program to find the answer. Since it can be extremely large, print the count modulo 10^9+7.

Constraints

* 1 \leq N \leq 3000
* 1 \leq M \leq 3000

Input

Input is given from Standard Input in the following format:


N M


Output

Print the count of the different possible sequences of colors of 2 \times M bricks that will be stacked, modulo 10^9+7.

Examples

Input

2 3


Output

56


Input

1000 10


Output

1048576


Input

1000 3000


Output

693347555

### Response

```cpp
#include<bits/stdc++.h>
#define ll long long
#define fo(i,x,y) for(int i=x;i<=y;i++)
#define fd(i,x,y) for(int i=x;i>=y;i--)
using namespace std;

const int maxn=3007,mo=1e9+7;

int n,m,Ans;
int f[maxn][maxn][2];

void Init(){
	scanf("%d%d",&n,&m);
}

void Plus(int &a,int b){
	(a+=b)%=mo;
}
void Solve(){
	f[0][0][1]=1;
	fo(i,1,n) f[0][i][0]=1;
	fo(i,0,m-1){
		fo(j,0,n)
			fo(k,0,1){
				if (j>0){
					Plus(f[i+1][j-1][k|(j-1==0)],f[i][j][k]);
					Plus(f[i+1][j][k|(j-1==0)],f[i][j][k]);
				}
				if (j<n){
					Plus(f[i+1][j+1][k],f[i][j][k]);
					Plus(f[i+1][j][k],f[i][j][k]);
				}
			}
	}
}

void Print(){
	fo(i,0,n) Plus(Ans,f[m][i][1]);
	printf("%d\n",Ans);
}

int main(){
	Init();
	Solve();
	Print();
	return 0;
}
```