#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int mm=1000000007;
int i,j,k,n,m,f[2][4101][2],now,pre,ans,t1,t2;
void upd(int x,int y,int v){
	f[now][x][y]=(f[now][x][y]+v)%mm;
}
int main(){
	scanf("%d%d",&n,&m);memset(f,0,sizeof(f));now=0;pre=1;
	for (i=0;i<=n;i++) f[now][i][0]=1;
	for (i=0;i<m;i++){
		now^=1;pre^=1;memset(f[now],0,sizeof(f[now]));
		for (j=0;j<=n;j++)
		 for (k=0;k<=1;k++){
		 	if (j){
		 		upd(j-1,(j==1)?1:k,f[pre][j][k]);
				upd(j,(j==1)?1:k,f[pre][j][k]);
			}
			if (j<n){
				upd(j+1,(j==0)?1:k,f[pre][j][k]);
				upd(j,(j==0)?1:k,f[pre][j][k]);
			}
		 }
	}
	for (i=ans=0;i<=n;i++) ans=(ans+f[now][i][1])%mm;
	printf("%d\n",ans);
	return 0;
}
### 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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int mm=1000000007;
int i,j,k,n,m,f[2][4101][2],now,pre,ans,t1,t2;
void upd(int x,int y,int v){
	f[now][x][y]=(f[now][x][y]+v)%mm;
}
int main(){
	scanf("%d%d",&n,&m);memset(f,0,sizeof(f));now=0;pre=1;
	for (i=0;i<=n;i++) f[now][i][0]=1;
	for (i=0;i<m;i++){
		now^=1;pre^=1;memset(f[now],0,sizeof(f[now]));
		for (j=0;j<=n;j++)
		 for (k=0;k<=1;k++){
		 	if (j){
		 		upd(j-1,(j==1)?1:k,f[pre][j][k]);
				upd(j,(j==1)?1:k,f[pre][j][k]);
			}
			if (j<n){
				upd(j+1,(j==0)?1:k,f[pre][j][k]);
				upd(j,(j==0)?1:k,f[pre][j][k]);
			}
		 }
	}
	for (i=ans=0;i<=n;i++) ans=(ans+f[now][i][1])%mm;
	printf("%d\n",ans);
	return 0;
}
```