#include<iostream>
#include<cstdio>
#include<cstring>

namespace dyshizhendeJM
{
	const int N=3030,MOD=1000000007;
	inline void inc(int a,int &b){b+=a;if(b>=MOD)b-=MOD;}

	int f[N][N][2];
	int n,m;

	void dp()
	{
		for(int i=0;i<=n;i++)f[0][i][i==0]=1;

		for(int i=0;i<m;i++)
			for(int j=0;j<=n;j++)
				for(int k=0;k<=1;k++)
				{
					int v=f[i][j][k];
					if(!v)continue;
					
					if(j>0)
					{
						inc(v,f[i+1][j][k|((j-1)==0)]);
						inc(v,f[i+1][j-1][k|((j-1)==0)]);
					}
					if(j<n)
					{
						inc(v,f[i+1][j][k]);
						inc(v,f[i+1][j+1][k]);
					}
				}
	}
	void solve()
	{
		scanf("%d%d",&n,&m);
		dp();
		int ans=0;
		for(int i=0;i<=n;i++)
			inc(f[m][i][1],ans);
		printf("%d\n",ans);

	}
}

int main()
{
	dyshizhendeJM::solve();
	return 0;
}

### 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<iostream>
#include<cstdio>
#include<cstring>

namespace dyshizhendeJM
{
	const int N=3030,MOD=1000000007;
	inline void inc(int a,int &b){b+=a;if(b>=MOD)b-=MOD;}

	int f[N][N][2];
	int n,m;

	void dp()
	{
		for(int i=0;i<=n;i++)f[0][i][i==0]=1;

		for(int i=0;i<m;i++)
			for(int j=0;j<=n;j++)
				for(int k=0;k<=1;k++)
				{
					int v=f[i][j][k];
					if(!v)continue;
					
					if(j>0)
					{
						inc(v,f[i+1][j][k|((j-1)==0)]);
						inc(v,f[i+1][j-1][k|((j-1)==0)]);
					}
					if(j<n)
					{
						inc(v,f[i+1][j][k]);
						inc(v,f[i+1][j+1][k]);
					}
				}
	}
	void solve()
	{
		scanf("%d%d",&n,&m);
		dp();
		int ans=0;
		for(int i=0;i<=n;i++)
			inc(f[m][i][1],ans);
		printf("%d\n",ans);

	}
}

int main()
{
	dyshizhendeJM::solve();
	return 0;
}

```