#include<bits/stdc++.h>
using namespace std;
#define ll long long
inline ll read() {
    char ch = getchar(); ll x = 0, f = 1;
    while(ch < '0' || ch > '9') {
        if(ch == '-') f = -1;
        ch = getchar();
    } while('0' <= ch && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    } return x * f;
}
const int kcz=1000000007;
#define jia(a,b) a+=b;if (a>=kcz) a-=kcz
const int maxn=3005;
int n,m,f[maxn][maxn][2];
int main(){
	n=read();m=read();f[0][0][1]=1;	
	for (int i=1;i<=n;i++) f[0][i][0]=1;
	for (int i=0;i<m;i++)
	for (int j=0;j<=n;j++)
	for (int k=0;k<=1;k++)
	if (f[i][j][k])
	{
		if (j) {jia(f[i+1][j-1][k|(j==1)],f[i][j][k]);}
		if (n-j) {jia(f[i+1][j+1][k],f[i][j][k]);}
		if (j){jia(f[i+1][j][k|(j==1)],f[i][j][k]);}
		if (n-j) {jia(f[i+1][j][k],f[i][j][k]);}
	}
	int ans=0;
	for (int i=0;i<=n;i++)
	{	jia(ans,f[m][i][1]);} 
	printf("%d\n",ans);
}
		
### Prompt

Construct a cpp code solution to the problem outlined:

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;
#define ll long long
inline ll read() {
    char ch = getchar(); ll x = 0, f = 1;
    while(ch < '0' || ch > '9') {
        if(ch == '-') f = -1;
        ch = getchar();
    } while('0' <= ch && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    } return x * f;
}
const int kcz=1000000007;
#define jia(a,b) a+=b;if (a>=kcz) a-=kcz
const int maxn=3005;
int n,m,f[maxn][maxn][2];
int main(){
	n=read();m=read();f[0][0][1]=1;	
	for (int i=1;i<=n;i++) f[0][i][0]=1;
	for (int i=0;i<m;i++)
	for (int j=0;j<=n;j++)
	for (int k=0;k<=1;k++)
	if (f[i][j][k])
	{
		if (j) {jia(f[i+1][j-1][k|(j==1)],f[i][j][k]);}
		if (n-j) {jia(f[i+1][j+1][k],f[i][j][k]);}
		if (j){jia(f[i+1][j][k|(j==1)],f[i][j][k]);}
		if (n-j) {jia(f[i+1][j][k],f[i][j][k]);}
	}
	int ans=0;
	for (int i=0;i<=n;i++)
	{	jia(ans,f[m][i][1]);} 
	printf("%d\n",ans);
}
		
```