#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 3e3 + 7;
const int MOD = 1e9 + 7;

int dp[N][N][2];

int main()
{
	int n, m;
	cin >> n >> m;
	for (int i=0; i<=n; i++)
		dp[0][i][1] = 1;
	for (int i=1; i<=m; i++)
	{
		for (int j=0; j<=n; j++)
		{
			if (j == 0)
			{
				dp[i][j][1] = dp[i][j][0] = (dp[i-1][j][1] + dp[i-1][j+1][1]) % MOD;
			}
			else if (j == n)
			{
				dp[i][j][1] = dp[i][j][0] = (dp[i-1][j][1] + dp[i-1][j-1][1]) % MOD;
			}
			else
				for (int k=0; k<2; k++)
					dp[i][j][k] = (dp[i-1][j][k] * 2LL + 1LL * dp[i-1][j-1][k] + 1LL * dp[i-1][j+1][k]) % MOD;
		}
	}
	int ans = 0;
	for (int i=0; i<=n; i++)
		ans = (ans + dp[m][i][0]) % MOD;
	cout << ans << "\n";
}
### 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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 3e3 + 7;
const int MOD = 1e9 + 7;

int dp[N][N][2];

int main()
{
	int n, m;
	cin >> n >> m;
	for (int i=0; i<=n; i++)
		dp[0][i][1] = 1;
	for (int i=1; i<=m; i++)
	{
		for (int j=0; j<=n; j++)
		{
			if (j == 0)
			{
				dp[i][j][1] = dp[i][j][0] = (dp[i-1][j][1] + dp[i-1][j+1][1]) % MOD;
			}
			else if (j == n)
			{
				dp[i][j][1] = dp[i][j][0] = (dp[i-1][j][1] + dp[i-1][j-1][1]) % MOD;
			}
			else
				for (int k=0; k<2; k++)
					dp[i][j][k] = (dp[i-1][j][k] * 2LL + 1LL * dp[i-1][j-1][k] + 1LL * dp[i-1][j+1][k]) % MOD;
		}
	}
	int ans = 0;
	for (int i=0; i<=n; i++)
		ans = (ans + dp[m][i][0]) % MOD;
	cout << ans << "\n";
}
```