#include<cstdio>
#define ll long long
#define mo 1000000007
using namespace std;
int f[3005][3005][4],p[6005],ans;
int n,m;
inline int get(int x){
	return !x?1:x==n?2:0;
}
inline void upd(int &x,int y){
	x=x+y;
	x=x>=mo?x-mo:x;
}
int main(){
	scanf("%d%d",&n,&m);
	for (int i=0;i<=n;i++)
		f[0][i][get(i)]=1;
	p[0]=1;
	for (int i=1;i<=2*m;i++) p[i]=p[i-1]*2%mo;
	ans=p[m*2];
	if (n<m){
		for (int i=1;i<=m;i++)
			for (int jzq=0;jzq<4;jzq++)
				for (int R=0;R<=n;R++){
					int B=n-R;
					if (R){
						upd(f[i][R-1][jzq|get(R-1)],f[i-1][R][jzq]);
					upd(f[i][R][jzq|get(R-1)],f[i-1][R][jzq]);
					}
					else if (jzq==3)
						upd(ans,-1ll*f[i-1][R][jzq]*p[2*(m-i)+1]%mo+mo);
					if (B){
						upd(f[i][R+1][jzq|get(R+1)],f[i-1][R][jzq]);
						upd(f[i][R][jzq|get(R+1)],f[i-1][R][jzq]);
					}
					else if (jzq==3)
						upd(ans,-1ll*f[i-1][R][jzq]*p[2*(m-i)+1]%mo+mo);
			}
	}
	printf("%lld\n",ans);
}

### Prompt

In CPP, your task is to solve 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<cstdio>
#define ll long long
#define mo 1000000007
using namespace std;
int f[3005][3005][4],p[6005],ans;
int n,m;
inline int get(int x){
	return !x?1:x==n?2:0;
}
inline void upd(int &x,int y){
	x=x+y;
	x=x>=mo?x-mo:x;
}
int main(){
	scanf("%d%d",&n,&m);
	for (int i=0;i<=n;i++)
		f[0][i][get(i)]=1;
	p[0]=1;
	for (int i=1;i<=2*m;i++) p[i]=p[i-1]*2%mo;
	ans=p[m*2];
	if (n<m){
		for (int i=1;i<=m;i++)
			for (int jzq=0;jzq<4;jzq++)
				for (int R=0;R<=n;R++){
					int B=n-R;
					if (R){
						upd(f[i][R-1][jzq|get(R-1)],f[i-1][R][jzq]);
					upd(f[i][R][jzq|get(R-1)],f[i-1][R][jzq]);
					}
					else if (jzq==3)
						upd(ans,-1ll*f[i-1][R][jzq]*p[2*(m-i)+1]%mo+mo);
					if (B){
						upd(f[i][R+1][jzq|get(R+1)],f[i-1][R][jzq]);
						upd(f[i][R][jzq|get(R+1)],f[i-1][R][jzq]);
					}
					else if (jzq==3)
						upd(ans,-1ll*f[i-1][R][jzq]*p[2*(m-i)+1]%mo+mo);
			}
	}
	printf("%lld\n",ans);
}

```