#include<bits/stdc++.h>
using namespace std;

#define MAX 3002

int n;
int m;

#define MOD 1000000007

long long int dp[MAX][MAX][2];
bool use[MAX][MAX][2];

inline long long int dfs(int a, int b, int c){
	if (b == 0){
		c = 1;
	}
	if (a == m){
		if (c == 1){
			return 1;
		}
		return 0;
	}
	if (use[a][b][c]){
		return dp[a][b][c];
	}
	use[a][b][c] = true;
	//up
	if (b != n){
		dp[a][b][c] += dfs(a + 1, b, c);
		dp[a][b][c] += dfs(a + 1, b + 1, c);
		dp[a][b][c] %= MOD;
	}
	if (b != 0){
		dp[a][b][c] += dfs(a + 1, b, c|(b==1));
		dp[a][b][c] += dfs(a + 1, b - 1, c|(b==1));
		dp[a][b][c] %= MOD;
	}
	return dp[a][b][c];
}
int main(){
	cin >> n >> m;
	long long int ans = 0;
	for (int i = 0; i <= n; i++){
		ans += dfs(0, i, 0);
		ans %= MOD;
	}
	printf("%lld\n", ans);
	return 0;
}
### Prompt

Create a solution in cpp for 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>
using namespace std;

#define MAX 3002

int n;
int m;

#define MOD 1000000007

long long int dp[MAX][MAX][2];
bool use[MAX][MAX][2];

inline long long int dfs(int a, int b, int c){
	if (b == 0){
		c = 1;
	}
	if (a == m){
		if (c == 1){
			return 1;
		}
		return 0;
	}
	if (use[a][b][c]){
		return dp[a][b][c];
	}
	use[a][b][c] = true;
	//up
	if (b != n){
		dp[a][b][c] += dfs(a + 1, b, c);
		dp[a][b][c] += dfs(a + 1, b + 1, c);
		dp[a][b][c] %= MOD;
	}
	if (b != 0){
		dp[a][b][c] += dfs(a + 1, b, c|(b==1));
		dp[a][b][c] += dfs(a + 1, b - 1, c|(b==1));
		dp[a][b][c] %= MOD;
	}
	return dp[a][b][c];
}
int main(){
	cin >> n >> m;
	long long int ans = 0;
	for (int i = 0; i <= n; i++){
		ans += dfs(0, i, 0);
		ans %= MOD;
	}
	printf("%lld\n", ans);
	return 0;
}
```