#include<bits/stdc++.h>
const int mo=1e9+7,N=3e3+5;
int n,m,i,j,ans,f[N][N];
inline void add(int&x,const int&y){x+=y-mo;x+=x>>31&mo;}
inline int calc(int n){
	if(n<1)return 0;
	memset(f,0,sizeof f);
	for(i=0;i<=n;++i)f[0][i]=1;
	for(i=0;i<m;++i)for(j=0;j<=n;++j){
		if(j)add(f[i+1][j],f[i][j]),add(f[i+1][j-1],f[i][j]);
		if(j<n)add(f[i+1][j],f[i][j]),add(f[i+1][j+1],f[i][j]);
	}
	ans=0;for(i=0;i<=n;++i)add(ans,f[m][i]);return ans;
}
int main(){
	scanf("%d%d",&n,&m);
	printf("%d\n",(calc(n)+mo-calc(n-1))%mo);
	return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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 mo=1e9+7,N=3e3+5;
int n,m,i,j,ans,f[N][N];
inline void add(int&x,const int&y){x+=y-mo;x+=x>>31&mo;}
inline int calc(int n){
	if(n<1)return 0;
	memset(f,0,sizeof f);
	for(i=0;i<=n;++i)f[0][i]=1;
	for(i=0;i<m;++i)for(j=0;j<=n;++j){
		if(j)add(f[i+1][j],f[i][j]),add(f[i+1][j-1],f[i][j]);
		if(j<n)add(f[i+1][j],f[i][j]),add(f[i+1][j+1],f[i][j]);
	}
	ans=0;for(i=0;i<=n;++i)add(ans,f[m][i]);return ans;
}
int main(){
	scanf("%d%d",&n,&m);
	printf("%d\n",(calc(n)+mo-calc(n-1))%mo);
	return 0;
}
```