#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>

#define fo(i,j,l) for(int i=j;i<=l;++i)
#define fd(i,j,l) for(int i=j;i>=l;--i)

using namespace std;
typedef long long ll;
const ll N=32e2,mo=1e9+7;

ll f[2][N][2];

int n,m;

int main()
{
	cin>>n>>m;
	fo(i,1,n)f[0][i][0]=1;
	f[0][0][1]=1;
	int u=0,v;
	fo(i,1,m){
		v=u^1;
		fo(i,0,n)f[v][i][0]=f[v][i][1]=0;
		fo(i,0,n)f[u][i][0]=f[u][i][0]%mo,f[u][i][1]=f[u][i][1]%mo;
		fo(i,1,n)f[v][i-1][(i==1)]=f[v][i-1][(i==1)]+f[u][i][0],f[v][i-1][1]=f[v][i-1][1]+f[u][i][1];
		fo(i,1,n)f[v][i][(i==1)]=f[v][i][(i==1)]+f[u][i][0],f[v][i][1]=f[v][i][1]+f[u][i][1];
		fo(i,0,n-1)f[v][i][0]=f[v][i][0]+f[u][i][0],f[v][i][1]=f[v][i][1]+f[u][i][1];
		fo(i,0,n-1)f[v][i+1][0]=f[v][i+1][0]+f[u][i][0],f[v][i+1][1]=f[v][i+1][1]+f[u][i][1];
		u=v;
	}
	ll ans=0;
	fo(i,0,n)ans=(ans+f[u][i][1])%mo;
	cout<<ans;
}
### Prompt

Your challenge is to write 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<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>

#define fo(i,j,l) for(int i=j;i<=l;++i)
#define fd(i,j,l) for(int i=j;i>=l;--i)

using namespace std;
typedef long long ll;
const ll N=32e2,mo=1e9+7;

ll f[2][N][2];

int n,m;

int main()
{
	cin>>n>>m;
	fo(i,1,n)f[0][i][0]=1;
	f[0][0][1]=1;
	int u=0,v;
	fo(i,1,m){
		v=u^1;
		fo(i,0,n)f[v][i][0]=f[v][i][1]=0;
		fo(i,0,n)f[u][i][0]=f[u][i][0]%mo,f[u][i][1]=f[u][i][1]%mo;
		fo(i,1,n)f[v][i-1][(i==1)]=f[v][i-1][(i==1)]+f[u][i][0],f[v][i-1][1]=f[v][i-1][1]+f[u][i][1];
		fo(i,1,n)f[v][i][(i==1)]=f[v][i][(i==1)]+f[u][i][0],f[v][i][1]=f[v][i][1]+f[u][i][1];
		fo(i,0,n-1)f[v][i][0]=f[v][i][0]+f[u][i][0],f[v][i][1]=f[v][i][1]+f[u][i][1];
		fo(i,0,n-1)f[v][i+1][0]=f[v][i+1][0]+f[u][i][0],f[v][i+1][1]=f[v][i+1][1]+f[u][i][1];
		u=v;
	}
	ll ans=0;
	fo(i,0,n)ans=(ans+f[u][i][1])%mo;
	cout<<ans;
}
```