#include<bits/stdc++.h>
using namespace std;
#define N 3333
#define mod 1000000007

int n,m,f[N][N][2],ans;

void pls(int &x,int y) {x=(x+y)%mod;}

int read(){
	int x=0,f=1;char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-f;
	for (;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
	return x*f;
}

int main(){
	n=read();m=read();
	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 k=0;k<=1;k++){
				if (j) pls(f[i+1][j-1][k|(j==1)],f[i][j][k]);
				if (j) pls(f[i+1][j][k|(j==1)],f[i][j][k]);
				if (n-j) pls(f[i+1][j][k],f[i][j][k]);
				if (n-j) pls(f[i+1][j+1][k],f[i][j][k]);
			}
	for (int i=0;i<=n;i++) pls(ans,f[m][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>
using namespace std;
#define N 3333
#define mod 1000000007

int n,m,f[N][N][2],ans;

void pls(int &x,int y) {x=(x+y)%mod;}

int read(){
	int x=0,f=1;char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-f;
	for (;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
	return x*f;
}

int main(){
	n=read();m=read();
	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 k=0;k<=1;k++){
				if (j) pls(f[i+1][j-1][k|(j==1)],f[i][j][k]);
				if (j) pls(f[i+1][j][k|(j==1)],f[i][j][k]);
				if (n-j) pls(f[i+1][j][k],f[i][j][k]);
				if (n-j) pls(f[i+1][j+1][k],f[i][j][k]);
			}
	for (int i=0;i<=n;i++) pls(ans,f[m][i][1]);
	printf("%d\n",ans);return 0;
}
```