#include<bits/stdc++.h>
#define N 3005
using namespace std;
const int mo=1e9+7;
int n,m,f[N][2],g[N][2],ans;
void inc(int &x,int y){ x=x+y>=mo ? x+y-mo : x+y;}
int main()
{
	int i,j;
	scanf("%d %d",&n,&m);
	for(i=1;i<=n;i++) f[i][0]=1;
	f[0][1]=1;
	while(m--){
		for(i=0;i<=n;i++) g[i][0]=f[i][0],g[i][1]=f[i][1],f[i][0]=f[i][1]=0;
		for(i=0;i<=n;i++)
			for(j=0;j<2;j++){
				if(i){
					inc(f[i][i==1 ? 1 : j],g[i][j]);
					inc(f[i-1][i==1 ? 1 : j],g[i][j]);
				  }
				if(i<n){
					inc(f[i][j],g[i][j]);
					inc(f[i+1][j],g[i][j]);
				  }
			  }
	  }
	for(i=0;i<=n;i++) inc(ans,f[i][1]);
	cout<<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>
#define N 3005
using namespace std;
const int mo=1e9+7;
int n,m,f[N][2],g[N][2],ans;
void inc(int &x,int y){ x=x+y>=mo ? x+y-mo : x+y;}
int main()
{
	int i,j;
	scanf("%d %d",&n,&m);
	for(i=1;i<=n;i++) f[i][0]=1;
	f[0][1]=1;
	while(m--){
		for(i=0;i<=n;i++) g[i][0]=f[i][0],g[i][1]=f[i][1],f[i][0]=f[i][1]=0;
		for(i=0;i<=n;i++)
			for(j=0;j<2;j++){
				if(i){
					inc(f[i][i==1 ? 1 : j],g[i][j]);
					inc(f[i-1][i==1 ? 1 : j],g[i][j]);
				  }
				if(i<n){
					inc(f[i][j],g[i][j]);
					inc(f[i+1][j],g[i][j]);
				  }
			  }
	  }
	for(i=0;i<=n;i++) inc(ans,f[i][1]);
	cout<<ans;
	return 0;
}

```