#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int md=1000000007;
int n,m,f[3005][3005][2],i,j,k;
int main()
{
	scanf("%d%d",&n,&m);
	for(i=1;i<=n;++i)
		f[0][i][0]=1;
	f[0][0][1]=1;
	for(i=0;i<m;++i)
		for(j=0;j<=n;++j)
			for(k=0;k<=1;++k)
				if(f[i][j][k])
				{
					if(j)
					{
						f[i+1][j][k||j==1]=(f[i+1][j][k||j==1]+f[i][j][k])%md;
						f[i+1][j-1][k||j==1]=(f[i+1][j-1][k||j==1]+f[i][j][k])%md;
					}
					if(j<n)
					{
						f[i+1][j][k]=(f[i+1][j][k]+f[i][j][k])%md;
						f[i+1][j+1][k]=(f[i+1][j+1][k]+f[i][j][k])%md;
					}
				}
	k=0;
	for(j=0;j<=n;++j)
		k=(k+f[m][j][1])%md;
	printf("%d",k);
	return 0;
}

### Prompt

In Cpp, your task is to solve 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<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int md=1000000007;
int n,m,f[3005][3005][2],i,j,k;
int main()
{
	scanf("%d%d",&n,&m);
	for(i=1;i<=n;++i)
		f[0][i][0]=1;
	f[0][0][1]=1;
	for(i=0;i<m;++i)
		for(j=0;j<=n;++j)
			for(k=0;k<=1;++k)
				if(f[i][j][k])
				{
					if(j)
					{
						f[i+1][j][k||j==1]=(f[i+1][j][k||j==1]+f[i][j][k])%md;
						f[i+1][j-1][k||j==1]=(f[i+1][j-1][k||j==1]+f[i][j][k])%md;
					}
					if(j<n)
					{
						f[i+1][j][k]=(f[i+1][j][k]+f[i][j][k])%md;
						f[i+1][j+1][k]=(f[i+1][j+1][k]+f[i][j][k])%md;
					}
				}
	k=0;
	for(j=0;j<=n;++j)
		k=(k+f[m][j][1])%md;
	printf("%d",k);
	return 0;
}

```