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
#include <cstdio>
#include <algorithm>

using namespace std;

const int N=3005,mo=1e9+7;

typedef long long LL;

int n,m,f[N][N][2],ans;

int main()
{
	scanf("%d%d",&n,&m);
	f[0][0][1]=1;
	for (int i=1;i<=n;i++) f[0][i][0]=1;
	for (int i=1,j,k;i<=m;i++)
	{
		for (j=0;j<=n;j++)
		{
			if (j>0)
			{
				k=(j==1)?1:0;
				f[i][j-1][1]=(f[i][j-1][1]+f[i-1][j][1])%mo;
				f[i][j][1]=(f[i][j][1]+f[i-1][j][1])%mo;
				f[i][j-1][k]=(f[i][j-1][k]+f[i-1][j][0])%mo;
				f[i][j][k]=(f[i][j][k]+f[i-1][j][0])%mo;
			}
			if (j<n)
			{
				for (k=0;k<2;k++)
				{
					f[i][j+1][k]=(f[i][j+1][k]+f[i-1][j][k])%mo;
					f[i][j][k]=(f[i][j][k]+f[i-1][j][k])%mo;
				}
			}
		}
	}
	for (int j=0;j<=n;j++) ans=(ans+f[m][j][1])%mo;
	printf("%d\n",ans);
	return 0;
}