#include<bits/stdc++.h>
const int N = 3e3 + 10;
using namespace std;

int f[N][N][2];
int n, m;
const int mod = 1e9 + 7;
inline void add(int &x,int y)
{  
   x += y; 
   if(x >= mod) x %= mod;
}

namespace solver
{
  void solve()
  {
    scanf("%d%d",&n,&m);
    for(int i=0; i<=m; ++i) f[0][i][!i]=1;
    for(int i=0; i<m; ++i)
	  for(int j=0; j<=n; ++j)
	    for(int v=0; v<2; ++v)
		{
		  int val=f[i][j][v];
		  if(j)
		  {
		    add(f[i+1][j-1][v|(j==1)], val);
			add(f[i+1][j][v|(j==1)], val);
		  }
		  if(n-j)
		  {
		    add(f[i+1][j][v], val);
			add(f[i+1][j+1][v], val);
		  }
		}
    int ans=0;
    for(int i=0; i<=n; ++i) add(ans,f[m][i][1]);
    printf("%d\n",ans);
  }
}

int main() { solver :: solve();}
### Prompt

Please formulate a cpp solution 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>
const int N = 3e3 + 10;
using namespace std;

int f[N][N][2];
int n, m;
const int mod = 1e9 + 7;
inline void add(int &x,int y)
{  
   x += y; 
   if(x >= mod) x %= mod;
}

namespace solver
{
  void solve()
  {
    scanf("%d%d",&n,&m);
    for(int i=0; i<=m; ++i) f[0][i][!i]=1;
    for(int i=0; i<m; ++i)
	  for(int j=0; j<=n; ++j)
	    for(int v=0; v<2; ++v)
		{
		  int val=f[i][j][v];
		  if(j)
		  {
		    add(f[i+1][j-1][v|(j==1)], val);
			add(f[i+1][j][v|(j==1)], val);
		  }
		  if(n-j)
		  {
		    add(f[i+1][j][v], val);
			add(f[i+1][j+1][v], val);
		  }
		}
    int ans=0;
    for(int i=0; i<=n; ++i) add(ans,f[m][i][1]);
    printf("%d\n",ans);
  }
}

int main() { solver :: solve();}
```