//start:
//finish:
#include<bits/stdc++.h>
#define MOD 1000000007
using namespace std;
typedef long long ll;
int n,m;
int f[3005][6005][2];
void add(int &x,int y){
	x=(x+y)%MOD;
}
int main(){
	scanf("%d%d",&n,&m);
	f[0][0][1]=1;
	for(int i=1;i<=n;i++)f[0][i][0]=1;
	for(int i=0;i<m;i++){
		for(int j=0;j<=n;j++){
			for(int t=0;t<=1;t++){
				if(!f[i][j][t])continue;
				int now=f[i][j][t];
				if(!j){
					add(f[i+1][j][1],now);
					add(f[i+1][j+1][1],now);
				}
				else if(j==n){
					add(f[i+1][j][t|(j==1)],now);
					add(f[i+1][j-1][t|(j==1)],now);
				}
				else{
					add(f[i+1][j][t],now);
					add(f[i+1][j+1][t],now);
					add(f[i+1][j][t|(j==1)],now);
					add(f[i+1][j-1][t|(j==1)],now);
				}
			}
		}
	}
	int ans=0;
	for(int i=0;i<=n;i++){
		add(ans,f[m][i][1]);
	}
	printf("%d\n",ans);
	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
//start:
//finish:
#include<bits/stdc++.h>
#define MOD 1000000007
using namespace std;
typedef long long ll;
int n,m;
int f[3005][6005][2];
void add(int &x,int y){
	x=(x+y)%MOD;
}
int main(){
	scanf("%d%d",&n,&m);
	f[0][0][1]=1;
	for(int i=1;i<=n;i++)f[0][i][0]=1;
	for(int i=0;i<m;i++){
		for(int j=0;j<=n;j++){
			for(int t=0;t<=1;t++){
				if(!f[i][j][t])continue;
				int now=f[i][j][t];
				if(!j){
					add(f[i+1][j][1],now);
					add(f[i+1][j+1][1],now);
				}
				else if(j==n){
					add(f[i+1][j][t|(j==1)],now);
					add(f[i+1][j-1][t|(j==1)],now);
				}
				else{
					add(f[i+1][j][t],now);
					add(f[i+1][j+1][t],now);
					add(f[i+1][j][t|(j==1)],now);
					add(f[i+1][j-1][t|(j==1)],now);
				}
			}
		}
	}
	int ans=0;
	for(int i=0;i<=n;i++){
		add(ans,f[m][i][1]);
	}
	printf("%d\n",ans);
	return 0;
}
```