#include <iostream>
#include <cstdio>

using namespace std;

const int P=1000000007;
const int N=3005;

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

inline void update(int &x,int y){(x+=y)%=P;}

int main()
{
	//freopen("piling.in","r",stdin),freopen("piling.out","w",stdout);
	scanf("%d%d",&n,&m);
	for (int i=0;i<=n;++i) update(f[0][i][i==0],1);
	for (int i=0,cur;i<m;++i)
		for (int j=0;j<=n;++j)
			for (int k=0;k<2;++k)
				if (cur=f[i][j][k])
				{
					if (j) update(f[i+1][j-1][k|(j==1)],cur),update(f[i+1][j][k|(j==1)],cur);
					if (n-j) update(f[i+1][j+1][k],cur),update(f[i+1][j][k],cur);
				}
	ans=0;
	for (int i=0;i<=n;++i) update(ans,f[m][i][1]);
	printf("%d\n",ans);
	//fclose(stdin),fclose(stdout);
	return 0;
}
### Prompt

Please create a solution in CPP 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 <iostream>
#include <cstdio>

using namespace std;

const int P=1000000007;
const int N=3005;

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

inline void update(int &x,int y){(x+=y)%=P;}

int main()
{
	//freopen("piling.in","r",stdin),freopen("piling.out","w",stdout);
	scanf("%d%d",&n,&m);
	for (int i=0;i<=n;++i) update(f[0][i][i==0],1);
	for (int i=0,cur;i<m;++i)
		for (int j=0;j<=n;++j)
			for (int k=0;k<2;++k)
				if (cur=f[i][j][k])
				{
					if (j) update(f[i+1][j-1][k|(j==1)],cur),update(f[i+1][j][k|(j==1)],cur);
					if (n-j) update(f[i+1][j+1][k],cur),update(f[i+1][j][k],cur);
				}
	ans=0;
	for (int i=0;i<=n;++i) update(ans,f[m][i][1]);
	printf("%d\n",ans);
	//fclose(stdin),fclose(stdout);
	return 0;
}
```