#include <bits/stdc++.h>

const int N=3010,Mod=1e9+7;

int n,m,dp[2][N][2];

void Init() {
	scanf("%d%d",&n,&m);
}

void Add(int &x,int v) { x+=v; if (x>=Mod) x-=Mod; }

void Solve() {
	for (int i=1;i<=n;++i) dp[0][i][0]=1; dp[0][0][1]=1;
	int op=0;
	for (int i=0;i<m;++i) {
		for (int j=0;j<=n;++j) for (int k=0;k<2;++k) dp[op^1][j][k]=0;
		for (int j=0;j<=n;++j) for (int k=0;k<2;++k) if (dp[op][j][k]) {
			if (j) Add(dp[op^1][j-1][(j-1)?k:1],dp[op][j][k]),Add(dp[op^1][j][(j-1)?k:1],dp[op][j][k]);
			if (n-j) Add(dp[op^1][j+1][k],dp[op][j][k]),Add(dp[op^1][j][k],dp[op][j][k]);
		}
		op^=1;
	}
	int ans=0;
	for (int i=0;i<=n;++i) ans=(ans+dp[op][i][1])%Mod;
	printf("%d\n",ans);
}

int main() {
	Init();
	Solve();
	return 0;
}
### Prompt

Please create a solution in Cpp 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>

const int N=3010,Mod=1e9+7;

int n,m,dp[2][N][2];

void Init() {
	scanf("%d%d",&n,&m);
}

void Add(int &x,int v) { x+=v; if (x>=Mod) x-=Mod; }

void Solve() {
	for (int i=1;i<=n;++i) dp[0][i][0]=1; dp[0][0][1]=1;
	int op=0;
	for (int i=0;i<m;++i) {
		for (int j=0;j<=n;++j) for (int k=0;k<2;++k) dp[op^1][j][k]=0;
		for (int j=0;j<=n;++j) for (int k=0;k<2;++k) if (dp[op][j][k]) {
			if (j) Add(dp[op^1][j-1][(j-1)?k:1],dp[op][j][k]),Add(dp[op^1][j][(j-1)?k:1],dp[op][j][k]);
			if (n-j) Add(dp[op^1][j+1][k],dp[op][j][k]),Add(dp[op^1][j][k],dp[op][j][k]);
		}
		op^=1;
	}
	int ans=0;
	for (int i=0;i<=n;++i) ans=(ans+dp[op][i][1])%Mod;
	printf("%d\n",ans);
}

int main() {
	Init();
	Solve();
	return 0;
}
```