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 <stdio.h>
#include <iostream>
#include <algorithm>
#include <memory.h>

using namespace std;
typedef long long LL;

const int mod = 1e9+7;
const int maxn = 3005;
int dp[maxn][maxn][2],n,m;
void add(int &u,int v) {
	u+=v;u>=mod?u-=mod:0;
}
int main()
{
	#ifdef Amberframe
		freopen("agc013d.in","r",stdin);
		freopen("agc013d.out","w",stdout);
	#endif
	scanf("%d %d",&n,&m);dp[0][0][1]=1;
	for (int i=1;i<=n;i++) dp[0][i][0]=1;
	for (int i=1;i<=m;i++)
	for (int j=0;j<=n;j++)
	for (int k=0;k<=1;k++)
	if (int w=dp[i-1][j][k]) {
		if (j>1) add(dp[i][j-1][k],w);
		if (j==1) add(dp[i][j-1][1],w);
		if (j<n) add(dp[i][j+1][k],w);
		if (j<n) add(dp[i][j][k],w);
		if (j>1) add(dp[i][j][k],w);
		if (j==1) add(dp[i][j][1],w);
	}
	int ans=0;
	for (int j=0;j<=n;j++)
		add(ans,dp[m][j][1]);
	printf("%d",ans);
	return 0;
}