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 <bits/stdc++.h>
using namespace std;

const int mo=int(1e9+7);
int n,m,f[3010][3010][2],ans;

void upd(int &x,int y){if ((x+=y)>=mo)  x-=mo;}

void work()
{
	scanf("%d %d",&n,&m);
	for (int i=1; i<=n; i++)  f[0][i][0]=1;
	f[0][0][1]=1;
	for (int i=0; i<m; i++)
		for (int j=0; j<=n; j++)
			for (int k=0; k<2; k++)
				if (f[i][j][k])
					{
						if (j>1)
							{
								upd(f[i+1][j-1][k],f[i][j][k]);
								upd(f[i+1][j][k],f[i][j][k]);
							}else  if (j==1)
							{
								upd(f[i+1][0][1],f[i][j][k]);
								upd(f[i+1][1][1],f[i][j][k]);
							}
						if (j<=n-1)
							{
								upd(f[i+1][j+1][k],f[i][j][k]);
								upd(f[i+1][j][k],f[i][j][k]);
							}
					}
	for (int i=0; i<=n; i++)  upd(ans,f[m][i][1]);
	printf("%d",ans);
}

int main()
{
	work();
	return 0;
}
