#include<bits/stdc++.h>
using namespace std;
#define mod 1000000007
int add(int x,int y){
	return (x+=y)<mod?x:x-mod;
}
int mul(int x,int y){
	return (int)((long long int)x*y%mod);
}
void addto(int&x,int y){
	x = add(x,y);
}
int d[3003][3003][2];
int main(){
	int n,m;
	scanf("%d%d",&n,&m);
	for(int i=0; i<=n; i++)
		d[0][i][0] = 1;
	for(int i=0; i<m; i++)
	for(int j=0; j<=n; j++)
	for(int k=0; k<2; k++){
		if(j){
			addto(d[i+1][j][k||(j==1)],d[i][j][k]);
			addto(d[i+1][j-1][k||(j==1)], d[i][j][k]);
		}
		if(j<n){
			addto(d[i+1][j][k||(j==0)],d[i][j][k]);
			addto(d[i+1][j+1][k||(j==0)],d[i][j][k]);
		}
	}
	int res = 0;
	for(int j=0; j<=n; j++)
		addto(res, d[m][j][1]);
	printf("%d\n",res);
	return 0;
}

### Prompt

Generate 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;
#define mod 1000000007
int add(int x,int y){
	return (x+=y)<mod?x:x-mod;
}
int mul(int x,int y){
	return (int)((long long int)x*y%mod);
}
void addto(int&x,int y){
	x = add(x,y);
}
int d[3003][3003][2];
int main(){
	int n,m;
	scanf("%d%d",&n,&m);
	for(int i=0; i<=n; i++)
		d[0][i][0] = 1;
	for(int i=0; i<m; i++)
	for(int j=0; j<=n; j++)
	for(int k=0; k<2; k++){
		if(j){
			addto(d[i+1][j][k||(j==1)],d[i][j][k]);
			addto(d[i+1][j-1][k||(j==1)], d[i][j][k]);
		}
		if(j<n){
			addto(d[i+1][j][k||(j==0)],d[i][j][k]);
			addto(d[i+1][j+1][k||(j==0)],d[i][j][k]);
		}
	}
	int res = 0;
	for(int j=0; j<=n; j++)
		addto(res, d[m][j][1]);
	printf("%d\n",res);
	return 0;
}

```