#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<deque>
#include<string>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#include<stdlib.h>
#include<cassert>
using namespace std;
const long long mod=1000000007;
const long long inf=mod*mod;
const long long d2=500000004;
const int M=3010;
long long dp[3100][3050][2];
int main(){
	int a,b;
	scanf("%d%d",&a,&b);
	for(int i=0;i<a;i++)dp[0][i][i==0]=1;
	for(int i=0;i<b-1;i++){
		for(int j=0;j<a;j++){
			for(int k=0;k<2;k++){
				if(!dp[i][j][k])continue;
 
				if(j)dp[i+1][j-1][k||(j==1)]=(dp[i+1][j-1][k||(j==1)]+dp[i][j][k])%mod;
				dp[i+1][j][k]=(dp[i+1][j][k]+dp[i][j][k]*2)%mod;
				if(j+1<a)dp[i+1][j+1][k]=(dp[i+1][j+1][k]+dp[i][j][k])%mod;
			}
		}
	}
	long long ret=0;
	for(int i=0;i<a;i++){
		ret=(ret+dp[b-1][i][1])%mod;
	}
	ret=ret*4%mod;
	printf("%lld\n",ret);
}
### 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<stdio.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<deque>
#include<string>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#include<stdlib.h>
#include<cassert>
using namespace std;
const long long mod=1000000007;
const long long inf=mod*mod;
const long long d2=500000004;
const int M=3010;
long long dp[3100][3050][2];
int main(){
	int a,b;
	scanf("%d%d",&a,&b);
	for(int i=0;i<a;i++)dp[0][i][i==0]=1;
	for(int i=0;i<b-1;i++){
		for(int j=0;j<a;j++){
			for(int k=0;k<2;k++){
				if(!dp[i][j][k])continue;
 
				if(j)dp[i+1][j-1][k||(j==1)]=(dp[i+1][j-1][k||(j==1)]+dp[i][j][k])%mod;
				dp[i+1][j][k]=(dp[i+1][j][k]+dp[i][j][k]*2)%mod;
				if(j+1<a)dp[i+1][j+1][k]=(dp[i+1][j+1][k]+dp[i][j][k])%mod;
			}
		}
	}
	long long ret=0;
	for(int i=0;i<a;i++){
		ret=(ret+dp[b-1][i][1])%mod;
	}
	ret=ret*4%mod;
	printf("%lld\n",ret);
}
```