#include <bits/stdc++.h>
#define rep(i,n) for ((i)=1;(i)<=(n);(i)++)
#define per(i,n) for ((i)=n;(i)>=(1);(i)--)
using namespace std;

const int mod=1e9+7;
int n,m,dp[3005][3005][2];
int i,j,k,ans;
void inc(int &x,int y){
	x=(x+y>=mod)?x+y-mod:x+y;
}
int main(){
	cin>>n>>m;
	rep(i,n) dp[0][i][0]=1;
	dp[0][0][1]=1;
	rep(i,m){
		for(j=0;j<=n;j++){
			for(k=0;k<2;k++){
				if(!dp[i-1][j][k]) continue;
				if(j){
					inc(dp[i][j-1][k||(j==1)],dp[i-1][j][k]);
				}
				if(j){
					inc(dp[i][j][k||(j==1)],dp[i-1][j][k]);
				}
				if(j<n){
					inc(dp[i][j][k],dp[i-1][j][k]);
				}
				if(j<n){
					inc(dp[i][j+1][k],dp[i-1][j][k]);
				}
			}
		}
	}
	for(i=0;i<=n;i++){
		inc(ans,dp[m][i][1]);
	}
	cout<<ans<<endl;
	return 0;
}
### Prompt

Please formulate 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 <bits/stdc++.h>
#define rep(i,n) for ((i)=1;(i)<=(n);(i)++)
#define per(i,n) for ((i)=n;(i)>=(1);(i)--)
using namespace std;

const int mod=1e9+7;
int n,m,dp[3005][3005][2];
int i,j,k,ans;
void inc(int &x,int y){
	x=(x+y>=mod)?x+y-mod:x+y;
}
int main(){
	cin>>n>>m;
	rep(i,n) dp[0][i][0]=1;
	dp[0][0][1]=1;
	rep(i,m){
		for(j=0;j<=n;j++){
			for(k=0;k<2;k++){
				if(!dp[i-1][j][k]) continue;
				if(j){
					inc(dp[i][j-1][k||(j==1)],dp[i-1][j][k]);
				}
				if(j){
					inc(dp[i][j][k||(j==1)],dp[i-1][j][k]);
				}
				if(j<n){
					inc(dp[i][j][k],dp[i-1][j][k]);
				}
				if(j<n){
					inc(dp[i][j+1][k],dp[i-1][j][k]);
				}
			}
		}
	}
	for(i=0;i<=n;i++){
		inc(ans,dp[m][i][1]);
	}
	cout<<ans<<endl;
	return 0;
}
```