#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int maxN=3010;
const int Mod=1e9+7;

int n,m;
int F[maxN][maxN][2];

void Plus(int &x,int y);
int main(){
    scanf("%d%d",&n,&m);
    F[0][0][1]=1;
    for (int i=1;i<=n;i++) F[0][i][0]=1;
    for (int i=0;i<m;i++) for (int j=0;j<=n;j++) for (int k=0;k<=1;k++){
        if (!F[i][j][k]) continue;
        if (j>=1) Plus(F[i+1][j-1][k|(j==1)],F[i][j][k]),Plus(F[i+1][j][k|(j==1)],F[i][j][k]);
        if (j<n) Plus(F[i+1][j][k],F[i][j][k]),Plus(F[i+1][j+1][k],F[i][j][k]);
    }
    int Ans=0;
    for (int i=0;i<=n;i++) Plus(Ans,F[m][i][1]);
    printf("%d\n",Ans);return 0;
}
void Plus(int &x,int y){
    x+=y;if (x>=Mod) x-=Mod;return;
}
### Prompt

Please create a solution in Cpp 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 maxN=3010;
const int Mod=1e9+7;

int n,m;
int F[maxN][maxN][2];

void Plus(int &x,int y);
int main(){
    scanf("%d%d",&n,&m);
    F[0][0][1]=1;
    for (int i=1;i<=n;i++) F[0][i][0]=1;
    for (int i=0;i<m;i++) for (int j=0;j<=n;j++) for (int k=0;k<=1;k++){
        if (!F[i][j][k]) continue;
        if (j>=1) Plus(F[i+1][j-1][k|(j==1)],F[i][j][k]),Plus(F[i+1][j][k|(j==1)],F[i][j][k]);
        if (j<n) Plus(F[i+1][j][k],F[i][j][k]),Plus(F[i+1][j+1][k],F[i][j][k]);
    }
    int Ans=0;
    for (int i=0;i<=n;i++) Plus(Ans,F[m][i][1]);
    printf("%d\n",Ans);return 0;
}
void Plus(int &x,int y){
    x+=y;if (x>=Mod) x-=Mod;return;
}
```