#include<cstdio>
#include<iostream>
using namespace std;
int n,m,ans;
int f[3010][3010][2];
int mod=1e9+7;
int plu(int &a,int b)
{
	a+=b;
	if (a>=mod) a-=mod;
}
int main()
{
	scanf("%d%d",&m,&n);
	for (int i=0;i<=m;i++)
		f[0][i][0]=1; 
	for (int i=1;i<=n;i++)
		for (int j=0;j<=m;j++)
		{
			int k=m-j;
			if (j)
			{
				plu(f[i][j][1],f[i-1][j][1]);
				plu(f[i][j][1],f[i-1][j-1][1]);
				if (j==1)
				{
					plu(f[i][j][1],f[i-1][j][0]);
					plu(f[i][j][1],f[i-1][j-1][0]);
				}
				else
				{
					plu(f[i][j][0],f[i-1][j][0]);
					plu(f[i][j][0],f[i-1][j-1][0]);
				}
			}
			if (k)
			{
				plu(f[i][j][1],f[i-1][j][1]);
				plu(f[i][j][1],f[i-1][j+1][1]);
				if (j==0)
				{
					plu(f[i][j][1],f[i-1][j][0]);
					plu(f[i][j][1],f[i-1][j+1][0]);
				}
				else
				{
					plu(f[i][j][0],f[i-1][j][0]);
					plu(f[i][j][0],f[i-1][j+1][0]);
				}	
			}
		}
	for (int i=0;i<=m;i++)
		plu(ans,f[n][i][1]);
	printf("%d",ans);
}
### Prompt

Generate a cpp solution to the following problem:

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>
#include<iostream>
using namespace std;
int n,m,ans;
int f[3010][3010][2];
int mod=1e9+7;
int plu(int &a,int b)
{
	a+=b;
	if (a>=mod) a-=mod;
}
int main()
{
	scanf("%d%d",&m,&n);
	for (int i=0;i<=m;i++)
		f[0][i][0]=1; 
	for (int i=1;i<=n;i++)
		for (int j=0;j<=m;j++)
		{
			int k=m-j;
			if (j)
			{
				plu(f[i][j][1],f[i-1][j][1]);
				plu(f[i][j][1],f[i-1][j-1][1]);
				if (j==1)
				{
					plu(f[i][j][1],f[i-1][j][0]);
					plu(f[i][j][1],f[i-1][j-1][0]);
				}
				else
				{
					plu(f[i][j][0],f[i-1][j][0]);
					plu(f[i][j][0],f[i-1][j-1][0]);
				}
			}
			if (k)
			{
				plu(f[i][j][1],f[i-1][j][1]);
				plu(f[i][j][1],f[i-1][j+1][1]);
				if (j==0)
				{
					plu(f[i][j][1],f[i-1][j][0]);
					plu(f[i][j][1],f[i-1][j+1][0]);
				}
				else
				{
					plu(f[i][j][0],f[i-1][j][0]);
					plu(f[i][j][0],f[i-1][j+1][0]);
				}	
			}
		}
	for (int i=0;i<=m;i++)
		plu(ans,f[n][i][1]);
	printf("%d",ans);
}
```