#include<bits/stdc++.h>
const int P=1e9+7;
int n,m,ans=0;
int f[6111][2];
int fix(int x){return x-=P,x+(x>>31&P);}
int main(){
	scanf("%d%d",&n,&m);
	n*=2;
	for(int i=0;i<=n;i+=2)f[i][!i]=1;
	for(int i=0;i<m;++i){
		for(int j=1;j<n;j+=2){
			f[j][0]=fix(f[j-1][0]+f[j+1][0]);
			f[j][1]=fix(f[j-1][1]+f[j+1][1]);
		}
		f[1][1]=(f[1][1]+f[1][0])%P;
		f[1][0]=0;
		for(int j=2;j<n;j+=2){
			f[j][0]=fix(f[j-1][0]+f[j+1][0]);
			f[j][1]=fix(f[j-1][1]+f[j+1][1]);
		}
		f[n][0]=f[n-1][0];
		f[n][1]=f[n-1][1];
		f[0][1]=f[1][1];
	}
	for(int i=0;i<=n;i+=2)ans=fix(ans+f[i][1]);
	printf("%d\n",ans);
	return 0;
}
### Prompt

Your task is to create 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>
const int P=1e9+7;
int n,m,ans=0;
int f[6111][2];
int fix(int x){return x-=P,x+(x>>31&P);}
int main(){
	scanf("%d%d",&n,&m);
	n*=2;
	for(int i=0;i<=n;i+=2)f[i][!i]=1;
	for(int i=0;i<m;++i){
		for(int j=1;j<n;j+=2){
			f[j][0]=fix(f[j-1][0]+f[j+1][0]);
			f[j][1]=fix(f[j-1][1]+f[j+1][1]);
		}
		f[1][1]=(f[1][1]+f[1][0])%P;
		f[1][0]=0;
		for(int j=2;j<n;j+=2){
			f[j][0]=fix(f[j-1][0]+f[j+1][0]);
			f[j][1]=fix(f[j-1][1]+f[j+1][1]);
		}
		f[n][0]=f[n-1][0];
		f[n][1]=f[n-1][1];
		f[0][1]=f[1][1];
	}
	for(int i=0;i<=n;i+=2)ans=fix(ans+f[i][1]);
	printf("%d\n",ans);
	return 0;
}
```