#include <bits/stdc++.h>
#define cal(a,b) ((a+=b)%=mo)
using namespace std;
const int N=6050,mo=1e9+7;
int gi(){
    char ch=getchar(); int x=0,q=0;
    while(ch<'0'||ch>'9') q=ch=='-'?1:q,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return q?-x:x;
}
int n,m,ans,f[2][N][2];
int main(){
    cin>>n>>m;
    for (int i=1; i<=n; ++i) f[0][i][0]=1;
    f[0][0][1]=1;
    int t=0;
    for (int i=0; i<2*m; ++i){
	t^=1; memset(f[t],0,sizeof(f[t]));
	for (int j=0; j<=2*m; ++j)
	    for (int p=0; p<=1; ++p)
		if (f[t^1][j][p]){
		    int z=j+(i&1);
		    if (n+2*(i/2)-j-i+(i&1)>=1)
			cal(f[t][z][p],f[t^1][j][p]);
		    if (z>=1)
			cal(f[t][z-1][p|(z==1)],f[t^1][j][p]);
		}
    }
    for (int i=0; i<=2*m; ++i)
	cal(ans,f[t][i][1]);
    cout<<ans;
    return 0;
}

### Prompt

Please formulate 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 cal(a,b) ((a+=b)%=mo)
using namespace std;
const int N=6050,mo=1e9+7;
int gi(){
    char ch=getchar(); int x=0,q=0;
    while(ch<'0'||ch>'9') q=ch=='-'?1:q,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return q?-x:x;
}
int n,m,ans,f[2][N][2];
int main(){
    cin>>n>>m;
    for (int i=1; i<=n; ++i) f[0][i][0]=1;
    f[0][0][1]=1;
    int t=0;
    for (int i=0; i<2*m; ++i){
	t^=1; memset(f[t],0,sizeof(f[t]));
	for (int j=0; j<=2*m; ++j)
	    for (int p=0; p<=1; ++p)
		if (f[t^1][j][p]){
		    int z=j+(i&1);
		    if (n+2*(i/2)-j-i+(i&1)>=1)
			cal(f[t][z][p],f[t^1][j][p]);
		    if (z>=1)
			cal(f[t][z-1][p|(z==1)],f[t^1][j][p]);
		}
    }
    for (int i=0; i<=2*m; ++i)
	cal(ans,f[t][i][1]);
    cout<<ans;
    return 0;
}

```