#include <bits/stdc++.h>
 
#define For(i, j, k) for(int i = j; i <= k; i++)
 
using namespace std;
 
const int N = 3010;
const int Mod = 1e9 + 7;
 
void add(int &x, int y){
	x += y;
	if(x >= Mod) x -= Mod;
}
 
int dp[N][N][2], n, m;
 
int main(){
	scanf("%d%d", &n, &m);
 
	For(i, 0, n) dp[0][i][i == 0] = 1;
	For(i, 0, m - 1) For(j, 0, n) For(v, 0, 1){
		int val = dp[i][j][v];
		if(j){
			add(dp[i + 1][j][v | (j == 1)], val);
			add(dp[i + 1][j - 1][v | (j == 1)], val);
		}
		if(n - j){
			add(dp[i + 1][j][v], val);
			add(dp[i + 1][j + 1][v], val);
		}
	}
 
	int ans = 0;
	For(i, 0, n) add(ans, dp[m][i][1]);
	printf("%d\n", ans);
	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>
 
#define For(i, j, k) for(int i = j; i <= k; i++)
 
using namespace std;
 
const int N = 3010;
const int Mod = 1e9 + 7;
 
void add(int &x, int y){
	x += y;
	if(x >= Mod) x -= Mod;
}
 
int dp[N][N][2], n, m;
 
int main(){
	scanf("%d%d", &n, &m);
 
	For(i, 0, n) dp[0][i][i == 0] = 1;
	For(i, 0, m - 1) For(j, 0, n) For(v, 0, 1){
		int val = dp[i][j][v];
		if(j){
			add(dp[i + 1][j][v | (j == 1)], val);
			add(dp[i + 1][j - 1][v | (j == 1)], val);
		}
		if(n - j){
			add(dp[i + 1][j][v], val);
			add(dp[i + 1][j + 1][v], val);
		}
	}
 
	int ans = 0;
	For(i, 0, n) add(ans, dp[m][i][1]);
	printf("%d\n", ans);
	return 0;
}
```