#include<bits/stdc++.h>
using namespace std;
const int N=3010,p=1e9+7;
void inc(int &x,int y){x+=y;if (x>=p) x-=p;}
int n,m,dp[N][N][2];
//dp[i][j][S]表示做了i次操作,剩下j个红球,红球是否恰好用完过的方案数
int main()
{
	scanf("%d%d",&m,&n);
	for (int i=1;i<=m;i++) dp[0][i][0]=1;
	dp[0][0][1]=1;
	for (int i=0;i<n;i++)
	for (int j=0;j<=m;j++)
	for (int k=0;k<2;k++){
		int v=dp[i][j][k];
		if (j){
			inc(dp[i+1][j][k|(j==1)],v);//RB
			inc(dp[i+1][j-1][k|(j==1)],v);//RR
		}
		if (j<m){
			inc(dp[i+1][j][k],v);//BR
			inc(dp[i+1][j+1][k],v);//BB
		}
	}
	int ans=0;
	for (int i=0;i<=m;i++) inc(ans,dp[n][i][1]);
	printf("%d\n",ans);
	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<bits/stdc++.h>
using namespace std;
const int N=3010,p=1e9+7;
void inc(int &x,int y){x+=y;if (x>=p) x-=p;}
int n,m,dp[N][N][2];
//dp[i][j][S]表示做了i次操作,剩下j个红球,红球是否恰好用完过的方案数
int main()
{
	scanf("%d%d",&m,&n);
	for (int i=1;i<=m;i++) dp[0][i][0]=1;
	dp[0][0][1]=1;
	for (int i=0;i<n;i++)
	for (int j=0;j<=m;j++)
	for (int k=0;k<2;k++){
		int v=dp[i][j][k];
		if (j){
			inc(dp[i+1][j][k|(j==1)],v);//RB
			inc(dp[i+1][j-1][k|(j==1)],v);//RR
		}
		if (j<m){
			inc(dp[i+1][j][k],v);//BR
			inc(dp[i+1][j+1][k],v);//BB
		}
	}
	int ans=0;
	for (int i=0;i<=m;i++) inc(ans,dp[n][i][1]);
	printf("%d\n",ans);
	return 0;
}
```