#include <bits/stdc++.h>
using namespace std;

inline int read()
{
	int x=0,f=1,c=getchar();
	while(c<48) c=='-'&&(f=-1),c=getchar();
	while(c>47) x=x*10+c-'0',c=getchar();
	return x*f;
}

const int MAXN = 5005;
const int mod = 1e9+7;
int f[MAXN][MAXN];
int n,m;

inline int dp(int x,int y,int ret=0)
{
	memset(f,0,sizeof(f));
	for(int i=0;i<=x;++i) f[0][i]=1;
	for(int i=1;i<=y;++i)
		for(int j=0;j<=x;++j)
		{
			if(j<x) (f[i][j]+=f[i-1][j+1])%=mod,(f[i][j]+=f[i-1][j])%=mod;
			if(j) (f[i][j]+=f[i-1][j])%=mod,(f[i][j]+=f[i-1][j-1])%=mod;
		}
	for(int i=0;i<=x;++i) (ret+=f[y][i])%=mod;
	return ret;
}

int main(int argc, char const *argv[])
{
	n=read(); m=read();
	printf("%d\n", (dp(n,m)-dp(n-1,m)+mod)%mod);
	return 0;
}


### Prompt

Your task is to create 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>
using namespace std;

inline int read()
{
	int x=0,f=1,c=getchar();
	while(c<48) c=='-'&&(f=-1),c=getchar();
	while(c>47) x=x*10+c-'0',c=getchar();
	return x*f;
}

const int MAXN = 5005;
const int mod = 1e9+7;
int f[MAXN][MAXN];
int n,m;

inline int dp(int x,int y,int ret=0)
{
	memset(f,0,sizeof(f));
	for(int i=0;i<=x;++i) f[0][i]=1;
	for(int i=1;i<=y;++i)
		for(int j=0;j<=x;++j)
		{
			if(j<x) (f[i][j]+=f[i-1][j+1])%=mod,(f[i][j]+=f[i-1][j])%=mod;
			if(j) (f[i][j]+=f[i-1][j])%=mod,(f[i][j]+=f[i-1][j-1])%=mod;
		}
	for(int i=0;i<=x;++i) (ret+=f[y][i])%=mod;
	return ret;
}

int main(int argc, char const *argv[])
{
	n=read(); m=read();
	printf("%d\n", (dp(n,m)-dp(n-1,m)+mod)%mod);
	return 0;
}


```