#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<stdlib.h>
#include<string.h>
#include<queue>
#define LL long long
using namespace std;
inline int read( )
{
  int sum=0;char c=getchar( );bool f=0;
  while(c<'0'||c>'9') {if(c=='-') f=1;c=getchar( );}
  while(c>='0'&&c<='9') {sum=sum*10+c-'0';c=getchar( );}
  if(f) return -sum;
  return sum;
}
const int mod=1000000007;
const int N=3005;
int n,m,f[N][N][2];
inline void up(int &x,int y) {x+=y;if(x>=mod) x-=mod;}
int main( )
{
  int i,j,k,x;
  n=read( );m=read( );
  for(i=0;i<=n;i++) f[0][i][0]=1;
  for(i=0;i<m;i++)
    for(j=0;j<=n;j++)
      for(k=0;k<=1;k++)
	if(x=f[i][j][k])
	  {
	    if(j!=n) up(f[i+1][j+1][k|(!j)],x),up(f[i+1][j][k|(!j)],x);
	    if(j) up(f[i+1][j-1][k|(j==1)],x),up(f[i+1][j][k|(j==1)],x);
	  }
  int ans=0;
  for(i=0;i<=n;i++) up(ans,f[m][i][1]);
  printf("%d",ans);
  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<stdio.h>
#include<iostream>
#include<algorithm>
#include<stdlib.h>
#include<string.h>
#include<queue>
#define LL long long
using namespace std;
inline int read( )
{
  int sum=0;char c=getchar( );bool f=0;
  while(c<'0'||c>'9') {if(c=='-') f=1;c=getchar( );}
  while(c>='0'&&c<='9') {sum=sum*10+c-'0';c=getchar( );}
  if(f) return -sum;
  return sum;
}
const int mod=1000000007;
const int N=3005;
int n,m,f[N][N][2];
inline void up(int &x,int y) {x+=y;if(x>=mod) x-=mod;}
int main( )
{
  int i,j,k,x;
  n=read( );m=read( );
  for(i=0;i<=n;i++) f[0][i][0]=1;
  for(i=0;i<m;i++)
    for(j=0;j<=n;j++)
      for(k=0;k<=1;k++)
	if(x=f[i][j][k])
	  {
	    if(j!=n) up(f[i+1][j+1][k|(!j)],x),up(f[i+1][j][k|(!j)],x);
	    if(j) up(f[i+1][j-1][k|(j==1)],x),up(f[i+1][j][k|(j==1)],x);
	  }
  int ans=0;
  for(i=0;i<=n;i++) up(ans,f[m][i][1]);
  printf("%d",ans);
  return 0;
}
	      
	   

```