#include<bits/stdc++.h>
using namespace std;
const int mod=1000000007,N=3100;
typedef long long ll;
inline void reduce(int&x){x+=x>>31&mod;}
int n,m,f[N][N][2];

int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n>>m;
	for(int i=0;i<=n;++i)f[0][i][!i]=1;
	for(int i=1;i<=m;++i)
		for(int j=0;j<=n;++j)
			for(int k=0;k<=1;++k)if(f[i-1][j][k]){
				int t=f[i-1][j][k]-mod;
				if(j)reduce(f[i][j][k|(j==1)]+=t),reduce(f[i][j-1][k|(j==1)]+=t);
				if(j!=n)reduce(f[i][j][k]+=t),reduce(f[i][j+1][k]+=t);
			}
	int ans=0;
	for(int i=0;i<=n;++i)reduce(ans+=f[m][i][1]-mod);
	cout<<ans<<'\n';
	return 0;
}

### Prompt

Please provide a cpp coded solution 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;
const int mod=1000000007,N=3100;
typedef long long ll;
inline void reduce(int&x){x+=x>>31&mod;}
int n,m,f[N][N][2];

int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n>>m;
	for(int i=0;i<=n;++i)f[0][i][!i]=1;
	for(int i=1;i<=m;++i)
		for(int j=0;j<=n;++j)
			for(int k=0;k<=1;++k)if(f[i-1][j][k]){
				int t=f[i-1][j][k]-mod;
				if(j)reduce(f[i][j][k|(j==1)]+=t),reduce(f[i][j-1][k|(j==1)]+=t);
				if(j!=n)reduce(f[i][j][k]+=t),reduce(f[i][j+1][k]+=t);
			}
	int ans=0;
	for(int i=0;i<=n;++i)reduce(ans+=f[m][i][1]-mod);
	cout<<ans<<'\n';
	return 0;
}

```