#include <cstdio>
#define MOD 1000000007
int dp[3005][3005][2];
int main()
{
	// freopen("AGC013-D.in", "r", stdin);
	int n, m;
	scanf("%d%d", &n, &m);
	for (int i = 0; i <= n; i++)
		dp[0][i][!i] = 1;
	for (int i = 0; i < m; i++)
	{
		for (int j = 0; j <= n; j++)
		{
			for (int x = 0; x < 2; x++)
			{
				if (j)
				{
					(dp[i + 1][j - 1][x | !(j - 1)] += dp[i][j][x]) %= MOD;
					(dp[i + 1][j][x | !(j - 1)] += dp[i][j][x]) %= MOD;
				}
				if (j != n)
				{
					(dp[i + 1][j + 1][x] += dp[i][j][x]) %= MOD;
					(dp[i + 1][j][x] += dp[i][j][x]) %= MOD;
				}
			}
		}
	}
	int ans = 0;
	for (int i = 0; i <= n; i++)
		(ans += dp[m][i][1]) %= MOD;
	printf("%d\n", ans);
	return 0;
}

### 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 <cstdio>
#define MOD 1000000007
int dp[3005][3005][2];
int main()
{
	// freopen("AGC013-D.in", "r", stdin);
	int n, m;
	scanf("%d%d", &n, &m);
	for (int i = 0; i <= n; i++)
		dp[0][i][!i] = 1;
	for (int i = 0; i < m; i++)
	{
		for (int j = 0; j <= n; j++)
		{
			for (int x = 0; x < 2; x++)
			{
				if (j)
				{
					(dp[i + 1][j - 1][x | !(j - 1)] += dp[i][j][x]) %= MOD;
					(dp[i + 1][j][x | !(j - 1)] += dp[i][j][x]) %= MOD;
				}
				if (j != n)
				{
					(dp[i + 1][j + 1][x] += dp[i][j][x]) %= MOD;
					(dp[i + 1][j][x] += dp[i][j][x]) %= MOD;
				}
			}
		}
	}
	int ans = 0;
	for (int i = 0; i <= n; i++)
		(ans += dp[m][i][1]) %= MOD;
	printf("%d\n", ans);
	return 0;
}

```