#include<bits/stdc++.h>
using namespace std;

int n,m,dp[2][3010][2],Ans;
const int mo=1e9+7;

int main()
{
#ifdef h10
	freopen("D.in","r",stdin);
	freopen("D.out","w",stdout);
#endif
	int i,fk=0;
	scanf("%d%d",&n,&m);
	dp[fk][0][1]=1;
	for (i=1;i<=n;i++)
		dp[fk][i][0]=1;
	while (m--)
	{
		memset(dp[fk^1],0,sizeof(dp[fk]));
		for (i=0;i<=n;i++)
		{
			if (i!=0)
			{
				(dp[fk^1][i-1][1]+=dp[fk][i][1])%=mo;
				(dp[fk^1][i-1][i==1]+=dp[fk][i][0])%=mo;
				(dp[fk^1][i][1]+=dp[fk][i][1])%=mo;
				(dp[fk^1][i][i==1]+=dp[fk][i][0])%=mo;
			}
			if (i!=n)
			{
				(dp[fk^1][i+1][1]+=dp[fk][i][1])%=mo;
				(dp[fk^1][i+1][0]+=dp[fk][i][0])%=mo;
				(dp[fk^1][i][1]+=dp[fk][i][1])%=mo;
				(dp[fk^1][i][0]+=dp[fk][i][0])%=mo;
			}
		}
		fk^=1;
	}
	for (i=0;i<=n;i++)
		(Ans+=dp[fk][i][1])%=mo;
	printf("%d\n",Ans);
}

### Prompt

Create a solution in cpp for 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;

int n,m,dp[2][3010][2],Ans;
const int mo=1e9+7;

int main()
{
#ifdef h10
	freopen("D.in","r",stdin);
	freopen("D.out","w",stdout);
#endif
	int i,fk=0;
	scanf("%d%d",&n,&m);
	dp[fk][0][1]=1;
	for (i=1;i<=n;i++)
		dp[fk][i][0]=1;
	while (m--)
	{
		memset(dp[fk^1],0,sizeof(dp[fk]));
		for (i=0;i<=n;i++)
		{
			if (i!=0)
			{
				(dp[fk^1][i-1][1]+=dp[fk][i][1])%=mo;
				(dp[fk^1][i-1][i==1]+=dp[fk][i][0])%=mo;
				(dp[fk^1][i][1]+=dp[fk][i][1])%=mo;
				(dp[fk^1][i][i==1]+=dp[fk][i][0])%=mo;
			}
			if (i!=n)
			{
				(dp[fk^1][i+1][1]+=dp[fk][i][1])%=mo;
				(dp[fk^1][i+1][0]+=dp[fk][i][0])%=mo;
				(dp[fk^1][i][1]+=dp[fk][i][1])%=mo;
				(dp[fk^1][i][0]+=dp[fk][i][0])%=mo;
			}
		}
		fk^=1;
	}
	for (i=0;i<=n;i++)
		(Ans+=dp[fk][i][1])%=mo;
	printf("%d\n",Ans);
}

```