#include <cstdio>
const int mo=1000000007;
int f[3005][3005][2];
int main(){
	int n,m;
	scanf("%d%d\n",&n,&m);
	for (int i=0;i<=n;i++) f[0][i][i==0]=1;
	for (int i=0;i<m;i++)
		for (int j=0;j<=n;j++)
			for (int k=0;k<2;k++)
			if (f[i][j][k]){
				int x=f[i][j][k];
				if (j) (f[i+1][j-1][k||j==1]+=x)%=mo,(f[i+1][j][k||j==1]+=x)%=mo;
				if (j<n) (f[i+1][j][k]+=x)%=mo,(f[i+1][j+1][k]+=x)%=mo;
			}
	int ans=0;
	for (int i=0;i<=n;i++) (ans+=f[m][i][1])%=mo;
	printf("%d\n",ans);
}
### Prompt

Create a solution in CPP for 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>
const int mo=1000000007;
int f[3005][3005][2];
int main(){
	int n,m;
	scanf("%d%d\n",&n,&m);
	for (int i=0;i<=n;i++) f[0][i][i==0]=1;
	for (int i=0;i<m;i++)
		for (int j=0;j<=n;j++)
			for (int k=0;k<2;k++)
			if (f[i][j][k]){
				int x=f[i][j][k];
				if (j) (f[i+1][j-1][k||j==1]+=x)%=mo,(f[i+1][j][k||j==1]+=x)%=mo;
				if (j<n) (f[i+1][j][k]+=x)%=mo,(f[i+1][j+1][k]+=x)%=mo;
			}
	int ans=0;
	for (int i=0;i<=n;i++) (ans+=f[m][i][1])%=mo;
	printf("%d\n",ans);
}
```