#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#define add(a,b) a=(a+b)%mod;
#define lol(i,n) for(ll i=0;i<n;i++)
#define mod 1000000007
typedef long long ll;

using namespace std;
#define N 3010
ll f[N][N],g[N][N],n,m;
int main(){
    lol(i,N)lol(j,N)f[i][j]=g[i][j]=0;
    cin>>n>>m;
    if(n==1){
	ll ans=2;
	lol(i,m)ans=ans*2%mod;
	cout<<ans<<endl;
	return 0;
    }
    for(ll i=1;i<=n;i++)f[0][i]=1; g[0][0]=1;
    for(ll i=0;i<m;i++){
	for(ll j=0;j<=n;j++){
	    for(ll k=j-1;k<=j+1;k++){
		if(k<0||k>n)continue;
		ll vf=((k==j&&j!=1&&j!=n)?2:1);
		ll vg=((k==j&&j!=0&&j!=n)?2:1);
		if(k>=1)add(f[i+1][k],f[i][j]*vf);
		if(k==0){
		    add(g[i+1][0],f[i][j]);
		    add(g[i+1][1],f[i][j]);
		}
		add(g[i+1][k],g[i][j]*vg);
	    }
	}
    }
    ll ans=0;
    for(ll i=0;i<=n;i++)add(ans,g[m][i]);
    cout<<ans<<endl;
    return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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<algorithm>
#include<vector>
#include<queue>
#define add(a,b) a=(a+b)%mod;
#define lol(i,n) for(ll i=0;i<n;i++)
#define mod 1000000007
typedef long long ll;

using namespace std;
#define N 3010
ll f[N][N],g[N][N],n,m;
int main(){
    lol(i,N)lol(j,N)f[i][j]=g[i][j]=0;
    cin>>n>>m;
    if(n==1){
	ll ans=2;
	lol(i,m)ans=ans*2%mod;
	cout<<ans<<endl;
	return 0;
    }
    for(ll i=1;i<=n;i++)f[0][i]=1; g[0][0]=1;
    for(ll i=0;i<m;i++){
	for(ll j=0;j<=n;j++){
	    for(ll k=j-1;k<=j+1;k++){
		if(k<0||k>n)continue;
		ll vf=((k==j&&j!=1&&j!=n)?2:1);
		ll vg=((k==j&&j!=0&&j!=n)?2:1);
		if(k>=1)add(f[i+1][k],f[i][j]*vf);
		if(k==0){
		    add(g[i+1][0],f[i][j]);
		    add(g[i+1][1],f[i][j]);
		}
		add(g[i+1][k],g[i][j]*vg);
	    }
	}
    }
    ll ans=0;
    for(ll i=0;i<=n;i++)add(ans,g[m][i]);
    cout<<ans<<endl;
    return 0;
}

```