#include<bits/stdc++.h>

using namespace std;
typedef long long ll;

const int N=3005,P=1000000007;
int n,m;
ll dp[N][N][2],ans;

int main(){
	scanf("%d%d",&n,&m);
	for(int i=0;i<=n;i++)dp[0][i][0]=1;
	for(int i=0;i<m;i++)for(int j=0;j<=n;j++)
		if(dp[i][j][0]||dp[i][j][1]){
			ll w0=dp[i][j][0]%P,w1=dp[i][j][1]%P;
			if(j>0){
				dp[i+1][j-1][1]+=w1,dp[i+1][j][1]+=w1;
				dp[i+1][j-1][j==1]+=w0,dp[i+1][j][j==1]+=w0;
			}
			if(j<n){
				dp[i+1][j+1][1]+=w1,dp[i+1][j][1]+=w1;
				dp[i+1][j+1][j==0]+=w0,dp[i+1][j][j==0]+=w0;
			}
		}
	for(int j=0;j<=n;j++) ans+=dp[m][j][1];
	printf("%lld\n",ans%P);
	return 0;
}
### Prompt

Please formulate 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;
typedef long long ll;

const int N=3005,P=1000000007;
int n,m;
ll dp[N][N][2],ans;

int main(){
	scanf("%d%d",&n,&m);
	for(int i=0;i<=n;i++)dp[0][i][0]=1;
	for(int i=0;i<m;i++)for(int j=0;j<=n;j++)
		if(dp[i][j][0]||dp[i][j][1]){
			ll w0=dp[i][j][0]%P,w1=dp[i][j][1]%P;
			if(j>0){
				dp[i+1][j-1][1]+=w1,dp[i+1][j][1]+=w1;
				dp[i+1][j-1][j==1]+=w0,dp[i+1][j][j==1]+=w0;
			}
			if(j<n){
				dp[i+1][j+1][1]+=w1,dp[i+1][j][1]+=w1;
				dp[i+1][j+1][j==0]+=w0,dp[i+1][j][j==0]+=w0;
			}
		}
	for(int j=0;j<=n;j++) ans+=dp[m][j][1];
	printf("%lld\n",ans%P);
	return 0;
}
```