#include<bits/stdc++.h>
using namespace std;
const int N=6005,mod=1000000007;
inline void add(int &a,int b){a+=b;if(a>=mod)a-=mod;}
int n,m,dp[2][N][2];
int main()
{
	scanf("%d%d",&n,&m);
	dp[0][0][1]=1;
	for(int i=1;i<=n;i++)
		dp[0][i][0]=1;
	int p=0;
	for(int i=0;i<2*m;i++)
	{
		memset(dp[p^1],0,sizeof(dp[p^1]));
		for(int j=0;j<=n;j++)
		{
			for(int s=0;s<2;s++)
			{
				int x=j,y=n-j+(i%2);
				if(!dp[p][j][s])
					continue;
				if(x) add(dp[p^1][j-1+(i+1)%2][s|(j==1)],dp[p][j][s]);
				if(y) add(dp[p^1][j+(i+1)%2][s],dp[p][j][s]);
			}
		}
		p^=1;
	}
	int ans=0;
	for(int i=0;i<=n;i++)
		add(ans,dp[p][i][1]);
	printf("%d\n",ans);
	return 0;
}
### 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<bits/stdc++.h>
using namespace std;
const int N=6005,mod=1000000007;
inline void add(int &a,int b){a+=b;if(a>=mod)a-=mod;}
int n,m,dp[2][N][2];
int main()
{
	scanf("%d%d",&n,&m);
	dp[0][0][1]=1;
	for(int i=1;i<=n;i++)
		dp[0][i][0]=1;
	int p=0;
	for(int i=0;i<2*m;i++)
	{
		memset(dp[p^1],0,sizeof(dp[p^1]));
		for(int j=0;j<=n;j++)
		{
			for(int s=0;s<2;s++)
			{
				int x=j,y=n-j+(i%2);
				if(!dp[p][j][s])
					continue;
				if(x) add(dp[p^1][j-1+(i+1)%2][s|(j==1)],dp[p][j][s]);
				if(y) add(dp[p^1][j+(i+1)%2][s],dp[p][j][s]);
			}
		}
		p^=1;
	}
	int ans=0;
	for(int i=0;i<=n;i++)
		add(ans,dp[p][i][1]);
	printf("%d\n",ans);
	return 0;
}
```