#include<cstdio>
#include<algorithm>
using u32=unsigned int;
const int P=1000000007;
u32 f[2][3007];
int cal(int m,int n)
{
    int now=0,sum=0;
    std::fill(f[0]+1,f[0]+m+1,1),f[0][0]=f[0][m+1]=f[1][0]=f[1][m+1]=0;
    for(int i=0;i<n;++i,now^=1) for(int j=1;j<=m;++j) f[!now][j]=(f[now][j-1]+2*f[now][j]+f[now][j+1])%P;
    for(int i=1;i<=m;++i) (sum+=f[now][i])%=P;
    return sum;
}
int main()
{
    int n,m;scanf("%d%d",&n,&m);
    printf("%d",4ll*(cal(n,m-1)-cal(n-1,m-1)+P)%P);
}

### 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<algorithm>
using u32=unsigned int;
const int P=1000000007;
u32 f[2][3007];
int cal(int m,int n)
{
    int now=0,sum=0;
    std::fill(f[0]+1,f[0]+m+1,1),f[0][0]=f[0][m+1]=f[1][0]=f[1][m+1]=0;
    for(int i=0;i<n;++i,now^=1) for(int j=1;j<=m;++j) f[!now][j]=(f[now][j-1]+2*f[now][j]+f[now][j+1])%P;
    for(int i=1;i<=m;++i) (sum+=f[now][i])%=P;
    return sum;
}
int main()
{
    int n,m;scanf("%d%d",&n,&m);
    printf("%d",4ll*(cal(n,m-1)-cal(n-1,m-1)+P)%P);
}

```