#include <cstdio>
#include <cstring>
#define MAXN 3010
#define LL long long

const LL P=1000000007;

int n,m;
LL f[MAXN][MAXN],f1[MAXN][MAXN];

inline void update(LL &x,LL y){ x=(x+y)%P; }

int main(){
#ifdef DEBUG
	freopen("D.in","r",stdin);
#endif
	scanf("%d%d",&n,&m);
	n--; m--;
	f[0][0]=1;
	for(int i=1;i<=m;i++)
		for(int j=0;j<=n;j++){
			update(f[i][j],f[i-1][j]*2);
			if(j>0) update(f[i][j-1],f[i-1][j]);
			if(j<n) update(f[i][j+1],f[i-1][j]);
		}
	f1[0][0]=1;
	for(int i=1;i<=m;i++)
		for(int j=0;j<=n;j++){
			if(j>0) update(f1[i][j],f1[i-1][j]*2);
			if(j>1) update(f1[i][j-1],f1[i-1][j]);
			if(j<n) update(f1[i][j+1],f1[i-1][j]);
		}
	LL ans=0;
	for(int i=0;i<=m;i++){
		LL t1=0,t2=0;
		for(int j=0;j<=n;j++){
			update(t1,f[i][j]);
			update(t2,f1[m-i][j]);
		}
		update(ans,t1*t2);
	}
	ans=ans*4%P;
	printf("%lld\n",ans);
	return 0;
}

### Prompt

Your challenge is to write 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 <cstdio>
#include <cstring>
#define MAXN 3010
#define LL long long

const LL P=1000000007;

int n,m;
LL f[MAXN][MAXN],f1[MAXN][MAXN];

inline void update(LL &x,LL y){ x=(x+y)%P; }

int main(){
#ifdef DEBUG
	freopen("D.in","r",stdin);
#endif
	scanf("%d%d",&n,&m);
	n--; m--;
	f[0][0]=1;
	for(int i=1;i<=m;i++)
		for(int j=0;j<=n;j++){
			update(f[i][j],f[i-1][j]*2);
			if(j>0) update(f[i][j-1],f[i-1][j]);
			if(j<n) update(f[i][j+1],f[i-1][j]);
		}
	f1[0][0]=1;
	for(int i=1;i<=m;i++)
		for(int j=0;j<=n;j++){
			if(j>0) update(f1[i][j],f1[i-1][j]*2);
			if(j>1) update(f1[i][j-1],f1[i-1][j]);
			if(j<n) update(f1[i][j+1],f1[i-1][j]);
		}
	LL ans=0;
	for(int i=0;i<=m;i++){
		LL t1=0,t2=0;
		for(int j=0;j<=n;j++){
			update(t1,f[i][j]);
			update(t2,f1[m-i][j]);
		}
		update(ans,t1*t2);
	}
	ans=ans*4%P;
	printf("%lld\n",ans);
	return 0;
}

```