#include <bits/stdc++.h>
#define mset(a,b) memset(a,b,sizeof a)
#define mcpy(a,b) memcpy(a,b,sizeof b)
#define lb(x) ((x)&(-(x)))
#define xx first
#define yy second
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define pii pair<int,int> 
#define dalao 1000000007
#define inf 0x3f3f3f3f
#define N 3005
using namespace std;
typedef long long ll;
int n,m,f[N][N][2];
inline void upd(int& a,int b){a=(a+b)%dalao;}
int main(){
	cin>>n>>m;
	for(int i=0;i<=n;i++)f[0][i][i==0]=1;
	for(int i=0;i<m;i++)for(int j=0;j<=n;j++)for(int k=0;k<2;k++)if(f[i][j][k]){
		int val=f[i][j][k];
		if(j>0){
			upd(f[i+1][j][k|(j==1)],val);
			upd(f[i+1][j-1][k|(j==1)],val);
		}
		if(j<n){
			upd(f[i+1][j][k],val);
			upd(f[i+1][j+1][k],val);
		}
	}
	int ans=0;
	for(int i=0;i<=n;i++)upd(ans,f[m][i][1]);
	cout<<ans;
	return 0;
}
### Prompt

Generate 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>
#define mset(a,b) memset(a,b,sizeof a)
#define mcpy(a,b) memcpy(a,b,sizeof b)
#define lb(x) ((x)&(-(x)))
#define xx first
#define yy second
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define pii pair<int,int> 
#define dalao 1000000007
#define inf 0x3f3f3f3f
#define N 3005
using namespace std;
typedef long long ll;
int n,m,f[N][N][2];
inline void upd(int& a,int b){a=(a+b)%dalao;}
int main(){
	cin>>n>>m;
	for(int i=0;i<=n;i++)f[0][i][i==0]=1;
	for(int i=0;i<m;i++)for(int j=0;j<=n;j++)for(int k=0;k<2;k++)if(f[i][j][k]){
		int val=f[i][j][k];
		if(j>0){
			upd(f[i+1][j][k|(j==1)],val);
			upd(f[i+1][j-1][k|(j==1)],val);
		}
		if(j<n){
			upd(f[i+1][j][k],val);
			upd(f[i+1][j+1][k],val);
		}
	}
	int ans=0;
	for(int i=0;i<=n;i++)upd(ans,f[m][i][1]);
	cout<<ans;
	return 0;
}
```