#include<bits/stdc++.h>

using namespace std;
const int maxn = 3005;
const int mod = 1e9+7;
void up(int &x,int y) { x+=y; if(x>=mod)x-=mod; }
int  f[maxn][maxn][2];
int n,m;
int main() {
	scanf("%d%d",&n,&m);
	f[0][0][1]=1;
	for(int i=1;i<=n;i++) f[0][i][0]=1;
	for(int i=1;i<=m;i++) {
		for(int j=0;j<=n;j++) {
			for(int k=0;k<=1;k++) {
				if(j) up(f[i][j][k|(j==1)],f[i-1][j][k]),up(f[i][j-1][k|(j==1)],f[i-1][j][k]);
				if(n-j) up(f[i][j+1][k],f[i-1][j][k]),up(f[i][j][k],f[i-1][j][k]);
			}
		}
	}
	int ans = 0;
	for(int i=0;i<=n;i++) up(ans,f[m][i][1]);
	printf("%d",ans);
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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;
const int maxn = 3005;
const int mod = 1e9+7;
void up(int &x,int y) { x+=y; if(x>=mod)x-=mod; }
int  f[maxn][maxn][2];
int n,m;
int main() {
	scanf("%d%d",&n,&m);
	f[0][0][1]=1;
	for(int i=1;i<=n;i++) f[0][i][0]=1;
	for(int i=1;i<=m;i++) {
		for(int j=0;j<=n;j++) {
			for(int k=0;k<=1;k++) {
				if(j) up(f[i][j][k|(j==1)],f[i-1][j][k]),up(f[i][j-1][k|(j==1)],f[i-1][j][k]);
				if(n-j) up(f[i][j+1][k],f[i-1][j][k]),up(f[i][j][k],f[i-1][j][k]);
			}
		}
	}
	int ans = 0;
	for(int i=0;i<=n;i++) up(ans,f[m][i][1]);
	printf("%d",ans);
}
```