#include<bits/stdc++.h>
#define mod 1000000007
#define N 3009
#define ad(x,y) (x=(x+(y))%mod)
using namespace std;

int n,m,dp[N][N];
int calc(int n,int m){
	int i,j; memset(dp,0,sizeof(dp));
	for (i=0; i<=n; i++) dp[0][i]=1;
	for (i=1; i<=m; i++)
		for (j=0; j<n; j++){
			dp[i][j]=(dp[i-1][j]<<1)%mod;
			if (j) ad(dp[i][j],dp[i-1][j-1]);
			if (j+1<n) ad(dp[i][j],dp[i-1][j+1]);
		}
	int ans=0;
	for (i=0; i<n; i++) ad(ans,dp[m][i]);
	return ans;
}
int main(){
	scanf("%d%d",&n,&m); m--;
	printf("%lld\n",4ll*(calc(n,m)-calc(n-1,m)+mod)%mod);
	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 mod 1000000007
#define N 3009
#define ad(x,y) (x=(x+(y))%mod)
using namespace std;

int n,m,dp[N][N];
int calc(int n,int m){
	int i,j; memset(dp,0,sizeof(dp));
	for (i=0; i<=n; i++) dp[0][i]=1;
	for (i=1; i<=m; i++)
		for (j=0; j<n; j++){
			dp[i][j]=(dp[i-1][j]<<1)%mod;
			if (j) ad(dp[i][j],dp[i-1][j-1]);
			if (j+1<n) ad(dp[i][j],dp[i-1][j+1]);
		}
	int ans=0;
	for (i=0; i<n; i++) ad(ans,dp[m][i]);
	return ans;
}
int main(){
	scanf("%d%d",&n,&m); m--;
	printf("%lld\n",4ll*(calc(n,m)-calc(n-1,m)+mod)%mod);
	return 0;
}

```