#include <iostream>
#include <algorithm>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<vector>
#include<iomanip>
#define sqr(x) (x)*(x)
using namespace std;
long long n,m,i,j,dp[3005][3005],llx=1e9+7,ans;
int main()
{
	cin>>n>>m;
	for (i=0;i<=n;i++) dp[0][i]=1;
	for (i=1;i<=m;i++)
	{
		for (j=0;j<=n;j++)
		{
			if (j>0) 
			{
				(dp[i][j]+=dp[i-1][j-1]+dp[i-1][j])%=llx;
			}
			if (j<n)
			{
				(dp[i][j]+=dp[i-1][j+1]+dp[i-1][j])%=llx;
			}
		}
	}
	for (i=0;i<=n;i++) (ans+=dp[m][i])%=llx;
	n--;
	memset(dp,0,sizeof(dp));
	for (i=0;i<=n;i++) dp[0][i]=1;
	for (i=1;i<=m;i++)
	{
		for (j=0;j<=n;j++)
		{
			if (j>0) 
			{
				(dp[i][j]+=dp[i-1][j-1]+dp[i-1][j])%=llx;
			}
			if (j<n)
			{
				(dp[i][j]+=dp[i-1][j+1]+dp[i-1][j])%=llx;
			}
		}
	}
	for (i=0;i<=n;i++) (ans-=dp[m][i])%=llx;
	cout<<(ans+llx)%llx;
	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<cmath>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<vector>
#include<iomanip>
#define sqr(x) (x)*(x)
using namespace std;
long long n,m,i,j,dp[3005][3005],llx=1e9+7,ans;
int main()
{
	cin>>n>>m;
	for (i=0;i<=n;i++) dp[0][i]=1;
	for (i=1;i<=m;i++)
	{
		for (j=0;j<=n;j++)
		{
			if (j>0) 
			{
				(dp[i][j]+=dp[i-1][j-1]+dp[i-1][j])%=llx;
			}
			if (j<n)
			{
				(dp[i][j]+=dp[i-1][j+1]+dp[i-1][j])%=llx;
			}
		}
	}
	for (i=0;i<=n;i++) (ans+=dp[m][i])%=llx;
	n--;
	memset(dp,0,sizeof(dp));
	for (i=0;i<=n;i++) dp[0][i]=1;
	for (i=1;i<=m;i++)
	{
		for (j=0;j<=n;j++)
		{
			if (j>0) 
			{
				(dp[i][j]+=dp[i-1][j-1]+dp[i-1][j])%=llx;
			}
			if (j<n)
			{
				(dp[i][j]+=dp[i-1][j+1]+dp[i-1][j])%=llx;
			}
		}
	}
	for (i=0;i<=n;i++) (ans-=dp[m][i])%=llx;
	cout<<(ans+llx)%llx;
	return 0;
}
```