#include<bits/stdc++.h>
#define L long long
using namespace std;
const int q=1000000007;
int n,m,f[3010][3010][2],p;
int main()
{
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	int i,j,k;
	scanf("%d%d",&n,&m);
	n--;
	m--;
	for(i=1;i<=n;i++)
	  f[0][i][0]=1;
	f[0][0][1]=1;
	for(i=1;i<=m;i++)
	  {
	   for(j=0;j<=n;j++)
	     for(k=0;k<=1;k++)
	       f[i][j][k]=(f[i-1][j][k]*2ll+f[i-1][j+1][k]+(j?f[i-1][j-1][k]:0))%q;
	   f[i][0][1]=(f[i][0][1]+f[i][0][0])%q;
	   f[i][0][0]=0;
      }
	for(i=0;i<=n;i++)
	  p=(p+f[m][i][1])%q;
	p=p*4ll%q;
	printf("%d\n",p);
	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<bits/stdc++.h>
#define L long long
using namespace std;
const int q=1000000007;
int n,m,f[3010][3010][2],p;
int main()
{
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	int i,j,k;
	scanf("%d%d",&n,&m);
	n--;
	m--;
	for(i=1;i<=n;i++)
	  f[0][i][0]=1;
	f[0][0][1]=1;
	for(i=1;i<=m;i++)
	  {
	   for(j=0;j<=n;j++)
	     for(k=0;k<=1;k++)
	       f[i][j][k]=(f[i-1][j][k]*2ll+f[i-1][j+1][k]+(j?f[i-1][j-1][k]:0))%q;
	   f[i][0][1]=(f[i][0][1]+f[i][0][0])%q;
	   f[i][0][0]=0;
      }
	for(i=0;i<=n;i++)
	  p=(p+f[m][i][1])%q;
	p=p*4ll%q;
	printf("%d\n",p);
	return 0;
}

```