#include <cstdio>
#include <cstring>
#define fo(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int N=3500,mo=1e9+7;
int m,n;
int f[N][N];
int DP(int n,int m)
{
    memset(f,0,sizeof(f));
    fo(i,0,n)f[0][i]=1;
    fo(i,1,m)fo(j,0,n)
    {
        if(j<n)((f[i][j]+=f[i-1][j+1])>=mo?f[i][j]-=mo:0),((f[i][j]+=f[i-1][j])>=mo?f[i][j]-=mo:0);
        if(j)((f[i][j]+=f[i-1][j])>=mo?f[i][j]-=mo:0),((f[i][j]+=f[i-1][j-1])>=mo?f[i][j]-=mo:0);
    }
    int ans=0;
    fo(i,0,n)((ans+=f[m][i])>=mo?ans-=mo:0);
    return ans;
}
int main()
{
    scanf("%d%d",&n,&m);
    printf("%d\n",(DP(n,m)-DP(n-1,m)+mo)%mo);
    return 0;
} 
### Prompt

In Cpp, your task is to solve 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>
#define fo(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int N=3500,mo=1e9+7;
int m,n;
int f[N][N];
int DP(int n,int m)
{
    memset(f,0,sizeof(f));
    fo(i,0,n)f[0][i]=1;
    fo(i,1,m)fo(j,0,n)
    {
        if(j<n)((f[i][j]+=f[i-1][j+1])>=mo?f[i][j]-=mo:0),((f[i][j]+=f[i-1][j])>=mo?f[i][j]-=mo:0);
        if(j)((f[i][j]+=f[i-1][j])>=mo?f[i][j]-=mo:0),((f[i][j]+=f[i-1][j-1])>=mo?f[i][j]-=mo:0);
    }
    int ans=0;
    fo(i,0,n)((ans+=f[m][i])>=mo?ans-=mo:0);
    return ans;
}
int main()
{
    scanf("%d%d",&n,&m);
    printf("%d\n",(DP(n,m)-DP(n-1,m)+mo)%mo);
    return 0;
} 
```