#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<queue>
#include<stack>
using namespace std;
#define MOD 1000000007
#define INF (1<<29)
#define LINF (1LL<<60)
#define EPS (1e-10)
typedef long long Int;
typedef pair<Int, Int> P;

Int res;
Int n, m;
int dp[6400][3240][2];
#define MID 3111
int main(){
  cin >> n >> m;
  for(int i = 0;i < 3240;i++)dp[0][i][!!i] = 1;
  for(int i = 0;i <= 2*m;i++){
dp[i][0][0] %= MOD;
    dp[i][0][0] += dp[i][0][1] %MOD;dp[i][0][0] %= MOD;
    dp[i][0][1] = 0;
    for(int j = 0;j <= n;j++){      
      if(i == 2*m)res = (res + dp[i][j][0]) % MOD;
      for(int k = 0;k < 2;k++){
	dp[i][j][k] %= MOD;
	if(i % 2 == 0){
	  if(j > 0)dp[i+1][j-1][k] += dp[i][j][k];
	  if(j < n)dp[i+1][j][k] += dp[i][j][k];
	}
	else{
	  dp[i+1][j][k] += dp[i][j][k];
	  dp[i+1][j+1][k] += dp[i][j][k];
	}
      }
    }
  }

  cout << res << endl;
}
### Prompt

Generate a Cpp solution to 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<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<queue>
#include<stack>
using namespace std;
#define MOD 1000000007
#define INF (1<<29)
#define LINF (1LL<<60)
#define EPS (1e-10)
typedef long long Int;
typedef pair<Int, Int> P;

Int res;
Int n, m;
int dp[6400][3240][2];
#define MID 3111
int main(){
  cin >> n >> m;
  for(int i = 0;i < 3240;i++)dp[0][i][!!i] = 1;
  for(int i = 0;i <= 2*m;i++){
dp[i][0][0] %= MOD;
    dp[i][0][0] += dp[i][0][1] %MOD;dp[i][0][0] %= MOD;
    dp[i][0][1] = 0;
    for(int j = 0;j <= n;j++){      
      if(i == 2*m)res = (res + dp[i][j][0]) % MOD;
      for(int k = 0;k < 2;k++){
	dp[i][j][k] %= MOD;
	if(i % 2 == 0){
	  if(j > 0)dp[i+1][j-1][k] += dp[i][j][k];
	  if(j < n)dp[i+1][j][k] += dp[i][j][k];
	}
	else{
	  dp[i+1][j][k] += dp[i][j][k];
	  dp[i+1][j+1][k] += dp[i][j][k];
	}
      }
    }
  }

  cout << res << endl;
}
```