#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<ctime>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
const int mod=1e9+7;
int dp[3001][3001][2];
int n,m;
int main()
{     int i,j,k,ans=0;
      scanf("%d%d",&n,&m);
      dp[0][0][1]=1;
      for(i=1;i<=n;i++)
         dp[0][i][0]=1;
      	for (i=0;i<m;i++)
	{
		for (j=0;j<=n;j++)
		{
			for (k=0;k<=1;k++)
			{
				if(j) dp[i+1][j-1][k|(j==1)]=(dp[i][j][k]+dp[i+1][j-1][k|(j==1)])%mod;
				if(j) dp[i+1][j][k|(j==1)]=(dp[i][j][k]+dp[i+1][j][k|(j==1)])%mod;
				if(n-j) dp[i+1][j][k]=(dp[i][j][k]+dp[i+1][j][k])%mod;
				if(n-j) dp[i+1][j+1][k]=(dp[i][j][k]+dp[i+1][j+1][k])%mod;
			}
		}
	}
	for (i=0;i<=n;i++)ans=(ans+dp[m][i][1])%mod;
	  cout<<ans%mod<<endl;
	  return 0;
}

### Prompt

Your task is to create 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<cstring>
#include<string>
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<ctime>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
const int mod=1e9+7;
int dp[3001][3001][2];
int n,m;
int main()
{     int i,j,k,ans=0;
      scanf("%d%d",&n,&m);
      dp[0][0][1]=1;
      for(i=1;i<=n;i++)
         dp[0][i][0]=1;
      	for (i=0;i<m;i++)
	{
		for (j=0;j<=n;j++)
		{
			for (k=0;k<=1;k++)
			{
				if(j) dp[i+1][j-1][k|(j==1)]=(dp[i][j][k]+dp[i+1][j-1][k|(j==1)])%mod;
				if(j) dp[i+1][j][k|(j==1)]=(dp[i][j][k]+dp[i+1][j][k|(j==1)])%mod;
				if(n-j) dp[i+1][j][k]=(dp[i][j][k]+dp[i+1][j][k])%mod;
				if(n-j) dp[i+1][j+1][k]=(dp[i][j][k]+dp[i+1][j+1][k])%mod;
			}
		}
	}
	for (i=0;i<=n;i++)ans=(ans+dp[m][i][1])%mod;
	  cout<<ans%mod<<endl;
	  return 0;
}

```