#include <bits/stdc++.h>
#define rep(i,n)for(int i=0;i<(n);i++)
#define MOD 1000000007
using namespace std;

int dp[5000][5000][2];
int main(){
	int n,m;cin>>n>>m;
	rep(i,n+1){
		if(i==0)dp[0][i][1]=1;
		else dp[0][i][0]=1;
	}
	rep(i,m)rep(j,n+1)rep(k,2){
		if(j>0){
			(dp[i+1][j-1][k||j==1]+=dp[i][j][k])%=MOD;
			(dp[i+1][j][k||j==1]+=dp[i][j][k])%=MOD;
		}
		if(j<n){
			(dp[i+1][j+1][k]+=dp[i][j][k])%=MOD;
			(dp[i+1][j][k]+=dp[i][j][k])%=MOD;
		}
	}
	int ans=0;
	rep(i,n+1)(ans+=dp[m][i][1])%=MOD;
	cout<<ans<<endl;
}
### Prompt

Please create a solution in Cpp 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>
#define rep(i,n)for(int i=0;i<(n);i++)
#define MOD 1000000007
using namespace std;

int dp[5000][5000][2];
int main(){
	int n,m;cin>>n>>m;
	rep(i,n+1){
		if(i==0)dp[0][i][1]=1;
		else dp[0][i][0]=1;
	}
	rep(i,m)rep(j,n+1)rep(k,2){
		if(j>0){
			(dp[i+1][j-1][k||j==1]+=dp[i][j][k])%=MOD;
			(dp[i+1][j][k||j==1]+=dp[i][j][k])%=MOD;
		}
		if(j<n){
			(dp[i+1][j+1][k]+=dp[i][j][k])%=MOD;
			(dp[i+1][j][k]+=dp[i][j][k])%=MOD;
		}
	}
	int ans=0;
	rep(i,n+1)(ans+=dp[m][i][1])%=MOD;
	cout<<ans<<endl;
}
```