#include<cstdio>
#include<cstring>
#include<algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,b,a) for(int i=b;i>=a;i--)
#define mset(a,x) memset(a,x,sizeof(a))
using namespace std;
typedef long long ll;
char ch;
int read(){int n=0,p=1;for(ch=getchar();ch<'0' || '9'<ch;ch=getchar())if(ch=='-') p=-1;for(;'0'<=ch && ch<='9';ch=getchar()) n=n*10+ch-'0';return n*p;}
const int N=3005,mo=1e9+7;
int n,m;
ll f[N][N],g[N][N];
void upd(ll &x,ll y){x=(x+y)%mo;}
int main()
{
	n=read(),m=read();
	g[0][0]=1;
	fo(i,1,n) f[0][i]=1;
	fo(i,1,m)
		fo(j,0,n)
		{
			upd(f[i][j],(j==n && n==1?0:f[i-1][j])+f[i-1][j+1]+(j>0?f[i-1][j-1]:0));
			upd(g[i][j],g[i-1][j]+(j>0?g[i-1][j-1]:0)+g[i-1][j+1]);
			if(0<j && j<n) upd(g[i][j],g[i-1][j]);
			if(1<j && j<n) upd(f[i][j],f[i-1][j]);
			if(j==1) upd(g[i][j],f[i-1][j]);
			if(j==0) upd(g[i][j],f[i][j]),f[i][0]=0;
		}
	ll ans=0;
	fo(i,0,n) ans=(ans+g[m][i])%mo;
	printf("%lld",ans);
	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<cstdio>
#include<cstring>
#include<algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,b,a) for(int i=b;i>=a;i--)
#define mset(a,x) memset(a,x,sizeof(a))
using namespace std;
typedef long long ll;
char ch;
int read(){int n=0,p=1;for(ch=getchar();ch<'0' || '9'<ch;ch=getchar())if(ch=='-') p=-1;for(;'0'<=ch && ch<='9';ch=getchar()) n=n*10+ch-'0';return n*p;}
const int N=3005,mo=1e9+7;
int n,m;
ll f[N][N],g[N][N];
void upd(ll &x,ll y){x=(x+y)%mo;}
int main()
{
	n=read(),m=read();
	g[0][0]=1;
	fo(i,1,n) f[0][i]=1;
	fo(i,1,m)
		fo(j,0,n)
		{
			upd(f[i][j],(j==n && n==1?0:f[i-1][j])+f[i-1][j+1]+(j>0?f[i-1][j-1]:0));
			upd(g[i][j],g[i-1][j]+(j>0?g[i-1][j-1]:0)+g[i-1][j+1]);
			if(0<j && j<n) upd(g[i][j],g[i-1][j]);
			if(1<j && j<n) upd(f[i][j],f[i-1][j]);
			if(j==1) upd(g[i][j],f[i-1][j]);
			if(j==0) upd(g[i][j],f[i][j]),f[i][0]=0;
		}
	ll ans=0;
	fo(i,0,n) ans=(ans+g[m][i])%mo;
	printf("%lld",ans);
	return 0;
}
```