#include <iostream>

using namespace std;

#define M 5005
#define MOD 1000000007
#define ll long long

int n, m;
ll ans, f[M][M][2];

int main()
{
	cin >> n >> m;
	for (int i = 1; i <= m; i++)
		f[0][i][0] = 1;
	f[0][0][1] = 1;

	for (int i = 0; i < m; i++)
		for (int j = 0; j <= n; j++)
		{
			if (j > 0)
			{
				if (j == 1) (f[i+1][j-1][1] += f[i][j][0]) %= MOD;
				else (f[i+1][j-1][0] += f[i][j][0]) %= MOD;
				(f[i+1][j-1][1] += f[i][j][1]) %= MOD;
				if (j == 1) (f[i+1][j][1] += f[i][j][0]) %= MOD;
				else (f[i+1][j][0] += f[i][j][0]) %= MOD;
				(f[i+1][j][1] += f[i][j][1]) %= MOD;
			}
			if (j < n)
			{
				(f[i+1][j+1][0] += f[i][j][0]) %= MOD;
				(f[i+1][j+1][1] += f[i][j][1]) %= MOD;
				(f[i+1][j][0] += f[i][j][0]) %= MOD;
				(f[i+1][j][1] += f[i][j][1]) %= MOD;
			}
		}

	for (int i = 0; i <= n; i++)
		(ans += f[m][i][1]) %= MOD;

	cout << ans;

	return 0;
}
### Prompt

Develop a solution in cpp 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 <iostream>

using namespace std;

#define M 5005
#define MOD 1000000007
#define ll long long

int n, m;
ll ans, f[M][M][2];

int main()
{
	cin >> n >> m;
	for (int i = 1; i <= m; i++)
		f[0][i][0] = 1;
	f[0][0][1] = 1;

	for (int i = 0; i < m; i++)
		for (int j = 0; j <= n; j++)
		{
			if (j > 0)
			{
				if (j == 1) (f[i+1][j-1][1] += f[i][j][0]) %= MOD;
				else (f[i+1][j-1][0] += f[i][j][0]) %= MOD;
				(f[i+1][j-1][1] += f[i][j][1]) %= MOD;
				if (j == 1) (f[i+1][j][1] += f[i][j][0]) %= MOD;
				else (f[i+1][j][0] += f[i][j][0]) %= MOD;
				(f[i+1][j][1] += f[i][j][1]) %= MOD;
			}
			if (j < n)
			{
				(f[i+1][j+1][0] += f[i][j][0]) %= MOD;
				(f[i+1][j+1][1] += f[i][j][1]) %= MOD;
				(f[i+1][j][0] += f[i][j][0]) %= MOD;
				(f[i+1][j][1] += f[i][j][1]) %= MOD;
			}
		}

	for (int i = 0; i <= n; i++)
		(ans += f[m][i][1]) %= MOD;

	cout << ans;

	return 0;
}
```