#include<bits/stdc++.h>
using namespace std;
#define N 3005
const int mod=1e9+7;
int n,m;
int dp[2][N][2];
int main(){
    scanf("%d%d",&n,&m);
    int now=0;
    for(int i=1;i<=n;i++) dp[now][i][0]=1;
    dp[now][0][1]=1;
    for(int i=1;i<=m;i++){
        memset(dp[now^1],0,sizeof(dp[0]));
        for(int j=0;j<=n;j++)
            for(int k=0;k<=1;k++){
                if(j!=0) 
                    (dp[now^1][j-1][(k==1)|(j==1)]+=dp[now][j][k])%=mod,
                    (dp[now^1][j][(k==1)|(j==1)]+=dp[now][j][k])%=mod;
                if(j!=n)
                    (dp[now^1][j][k]+=dp[now][j][k])%=mod,
                    (dp[now^1][j+1][k]+=dp[now][j][k])%=mod;
            }
        now^=1;
    }
    long long ans=0;
    for(int i=0;i<=n;i++)
        ans+=dp[now][i][1];
    cout<<ans%mod<<endl;
    return 0;
}
### Prompt

Develop a solution in cpp to the problem described below:

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<bits/stdc++.h>
using namespace std;
#define N 3005
const int mod=1e9+7;
int n,m;
int dp[2][N][2];
int main(){
    scanf("%d%d",&n,&m);
    int now=0;
    for(int i=1;i<=n;i++) dp[now][i][0]=1;
    dp[now][0][1]=1;
    for(int i=1;i<=m;i++){
        memset(dp[now^1],0,sizeof(dp[0]));
        for(int j=0;j<=n;j++)
            for(int k=0;k<=1;k++){
                if(j!=0) 
                    (dp[now^1][j-1][(k==1)|(j==1)]+=dp[now][j][k])%=mod,
                    (dp[now^1][j][(k==1)|(j==1)]+=dp[now][j][k])%=mod;
                if(j!=n)
                    (dp[now^1][j][k]+=dp[now][j][k])%=mod,
                    (dp[now^1][j+1][k]+=dp[now][j][k])%=mod;
            }
        now^=1;
    }
    long long ans=0;
    for(int i=0;i<=n;i++)
        ans+=dp[now][i][1];
    cout<<ans%mod<<endl;
    return 0;
}
```