#include<bits/stdc++.h>
#define N 3010
using namespace std;
const int MOD=1e9+7;
long long dp[N][N][2];
int n,m;
int ans;
int main(){
	scanf("%d%d",&n,&m);
	for(int i=0;i<n;i++) dp[0][i][0]=1;
	dp[0][n][1]=1;
	for(int i=1;i<=m;i++){
		for(int j=0;j<=n;j++){
			for(int l=(j==n?1:0);l<2;l++){
				if(j==0)       dp[i][j][l]=dp[i-1][j][l]  +dp[i-1][j+1][l];
				else if(j==n)  dp[i][j][1]=dp[i-1][j][1]  +dp[i-1][j-1][1]+dp[i-1][j-1][0];
				else           dp[i][j][l]=dp[i-1][j-1][l]+dp[i-1][j+1][l]+dp[i-1][j][l]*2;
				if(j==n-1)     dp[i][j][l]+=(l*2-1)*dp[i-1][j][0];
				dp[i][j][l]%=MOD;
			}
			ans+=dp[m][j][1],ans%=MOD;
		}
	}
	printf("%d\n",ans);
	return 0;
}
### Prompt

Create a solution in cpp for 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>
#define N 3010
using namespace std;
const int MOD=1e9+7;
long long dp[N][N][2];
int n,m;
int ans;
int main(){
	scanf("%d%d",&n,&m);
	for(int i=0;i<n;i++) dp[0][i][0]=1;
	dp[0][n][1]=1;
	for(int i=1;i<=m;i++){
		for(int j=0;j<=n;j++){
			for(int l=(j==n?1:0);l<2;l++){
				if(j==0)       dp[i][j][l]=dp[i-1][j][l]  +dp[i-1][j+1][l];
				else if(j==n)  dp[i][j][1]=dp[i-1][j][1]  +dp[i-1][j-1][1]+dp[i-1][j-1][0];
				else           dp[i][j][l]=dp[i-1][j-1][l]+dp[i-1][j+1][l]+dp[i-1][j][l]*2;
				if(j==n-1)     dp[i][j][l]+=(l*2-1)*dp[i-1][j][0];
				dp[i][j][l]%=MOD;
			}
			ans+=dp[m][j][1],ans%=MOD;
		}
	}
	printf("%d\n",ans);
	return 0;
}
```