#include<bits/stdc++.h>
#define ll long long
#define N 100005
#define mod 1000000007
#define For(i,x,y) for(int i=x;i<=y;i++)
#define Rep(i,x,y) for(int i=x;i>=y;i--)
using namespace std;
inline int read(){
	int x=0,f=0;char ch=getchar();
	for(;ch<'0'||ch>'9';ch=getchar()) if(ch=='-') f=1;
	for(;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-48;
	return f?-x:x;
}
int n,m;
ll f[3005][3005][2];
int main(){
	n=read();m=read()-2;
	For(i,0,n-1) f[0][i][i==0]=1;
	For(i,0,m) For(j,0,n) For(k,0,1) if(f[i][j][k]>0){
		if(j) f[i+1][j-1][(k)||j-1==0]=(f[i+1][j-1][(k)||j-1==0]+f[i][j][k])%mod;
		if(j<n-1) f[i+1][j+1][k]=(f[i+1][j+1][k]+f[i][j][k])%mod;
		f[i+1][j][k]=(f[i+1][j][k]+f[i][j][k]*2)%mod;
	}
	ll ans=0;
	For(i,0,n-1) ans=(ans+f[m+1][i][1])%mod;
	printf("%d\n",ans*4%mod);
	return 0;
}

### Prompt

Develop a solution in CPP 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<bits/stdc++.h>
#define ll long long
#define N 100005
#define mod 1000000007
#define For(i,x,y) for(int i=x;i<=y;i++)
#define Rep(i,x,y) for(int i=x;i>=y;i--)
using namespace std;
inline int read(){
	int x=0,f=0;char ch=getchar();
	for(;ch<'0'||ch>'9';ch=getchar()) if(ch=='-') f=1;
	for(;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-48;
	return f?-x:x;
}
int n,m;
ll f[3005][3005][2];
int main(){
	n=read();m=read()-2;
	For(i,0,n-1) f[0][i][i==0]=1;
	For(i,0,m) For(j,0,n) For(k,0,1) if(f[i][j][k]>0){
		if(j) f[i+1][j-1][(k)||j-1==0]=(f[i+1][j-1][(k)||j-1==0]+f[i][j][k])%mod;
		if(j<n-1) f[i+1][j+1][k]=(f[i+1][j+1][k]+f[i][j][k])%mod;
		f[i+1][j][k]=(f[i+1][j][k]+f[i][j][k]*2)%mod;
	}
	ll ans=0;
	For(i,0,n-1) ans=(ans+f[m+1][i][1])%mod;
	printf("%d\n",ans*4%mod);
	return 0;
}

```