#include <cstdio>
#define rep(i,j,k) for (i=j;i<=k;i++)
using namespace std;
const int N=3e3+5,mod=1e9+7;
int n,m,i,j,p,ans,f[N][N][2];
void updata(int &x,int y) { x+=y; if (x>=mod) x-=mod; }
int main()
{
//	freopen("piling.in","r",stdin);
//	freopen("piling.out","w",stdout);
	scanf("%d%d",&n,&m);
	f[0][0][1]=1;
	rep(i,1,n) f[0][i][0]=1;
	rep(i,0,m-1)
		rep(j,0,n)
			rep(p,0,1)
			if (f[i][j][p])
			{
				if (j) updata(f[i+1][j-1][p|(j==1)],f[i][j][p]);
				if (j) updata(f[i+1][j][p|(j==1)],f[i][j][p]);
				if (n-j) updata(f[i+1][j][p],f[i][j][p]);
				if (n-j) updata(f[i+1][j+1][p],f[i][j][p]);
			}
	rep(i,0,n)
		updata(ans,f[m][i][1]);
	printf("%d\n",ans);
	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 <cstdio>
#define rep(i,j,k) for (i=j;i<=k;i++)
using namespace std;
const int N=3e3+5,mod=1e9+7;
int n,m,i,j,p,ans,f[N][N][2];
void updata(int &x,int y) { x+=y; if (x>=mod) x-=mod; }
int main()
{
//	freopen("piling.in","r",stdin);
//	freopen("piling.out","w",stdout);
	scanf("%d%d",&n,&m);
	f[0][0][1]=1;
	rep(i,1,n) f[0][i][0]=1;
	rep(i,0,m-1)
		rep(j,0,n)
			rep(p,0,1)
			if (f[i][j][p])
			{
				if (j) updata(f[i+1][j-1][p|(j==1)],f[i][j][p]);
				if (j) updata(f[i+1][j][p|(j==1)],f[i][j][p]);
				if (n-j) updata(f[i+1][j][p],f[i][j][p]);
				if (n-j) updata(f[i+1][j+1][p],f[i][j][p]);
			}
	rep(i,0,n)
		updata(ans,f[m][i][1]);
	printf("%d\n",ans);
	return 0;
}
```