#include <cstdio>
#include <cstring>
#include <algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

const int N=3005,Mo=1e9+7;

void inc(int &x,int y) {
	x+=y;
	x-=(x>=Mo)?Mo:0; 
}

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

int Dp(int n,int m) {
	fo(i,0,m) fo(j,0,n) f[i][j]=0;
	fo(i,0,n) f[0][i]=1;
	fo(i,0,m-1)
		fo(j,0,n)
			if (f[i][j]) {	
				if (j>0) inc(f[i+1][j-1],f[i][j]);
				if (j>0) inc(f[i+1][j],f[i][j]);
				if (j<n) inc(f[i+1][j],f[i][j]);
				if (j<n) inc(f[i+1][j+1],f[i][j]); 	
			}
	int ans=0;
	fo(i,0,n) inc(ans,f[m][i]);
	return ans;
}

int main() {
	scanf("%d%d",&n,&m);
	printf("%d\n",(Dp(n,m)-Dp(n-1,m)+Mo)%Mo);
	return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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>
#include <algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

const int N=3005,Mo=1e9+7;

void inc(int &x,int y) {
	x+=y;
	x-=(x>=Mo)?Mo:0; 
}

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

int Dp(int n,int m) {
	fo(i,0,m) fo(j,0,n) f[i][j]=0;
	fo(i,0,n) f[0][i]=1;
	fo(i,0,m-1)
		fo(j,0,n)
			if (f[i][j]) {	
				if (j>0) inc(f[i+1][j-1],f[i][j]);
				if (j>0) inc(f[i+1][j],f[i][j]);
				if (j<n) inc(f[i+1][j],f[i][j]);
				if (j<n) inc(f[i+1][j+1],f[i][j]); 	
			}
	int ans=0;
	fo(i,0,n) inc(ans,f[m][i]);
	return ans;
}

int main() {
	scanf("%d%d",&n,&m);
	printf("%d\n",(Dp(n,m)-Dp(n-1,m)+Mo)%Mo);
	return 0;
}
```