# include <bits/stdc++.h>

using namespace std;
const int maxn = 3010;
const int mod = 1e9 + 7;

int f[maxn][maxn * 2];
int g[maxn][maxn * 2];
int n, m;

int main() {
	scanf("%d%d", &n, &m);
	--m; --n;
	g[0][0] = 1;
	for(int j = 1; j <= n; ++j) f[0][j] = 1;
	for(int i = 1; i <= m; ++i) {
		g[i][0] = (2ll * g[i-1][0] + g[i-1][1] + f[i-1][1]) % mod;
		for(int j = 1; j <= n; ++j) {
			f[i][j] = (2ll * f[i-1][j] + f[i-1][j-1] + f[i-1][j+1]) % mod;
			g[i][j] = (2ll * g[i-1][j] + g[i-1][j-1] + g[i-1][j+1]) % mod;
		}
	}
	int ans = 0;
	for(int j = 0; j <= n; ++j) ans = (ans + g[m][j]) % mod;
	ans = (ans << 1) % mod;
	ans = (ans << 1) % mod;
	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>

using namespace std;
const int maxn = 3010;
const int mod = 1e9 + 7;

int f[maxn][maxn * 2];
int g[maxn][maxn * 2];
int n, m;

int main() {
	scanf("%d%d", &n, &m);
	--m; --n;
	g[0][0] = 1;
	for(int j = 1; j <= n; ++j) f[0][j] = 1;
	for(int i = 1; i <= m; ++i) {
		g[i][0] = (2ll * g[i-1][0] + g[i-1][1] + f[i-1][1]) % mod;
		for(int j = 1; j <= n; ++j) {
			f[i][j] = (2ll * f[i-1][j] + f[i-1][j-1] + f[i-1][j+1]) % mod;
			g[i][j] = (2ll * g[i-1][j] + g[i-1][j-1] + g[i-1][j+1]) % mod;
		}
	}
	int ans = 0;
	for(int j = 0; j <= n; ++j) ans = (ans + g[m][j]) % mod;
	ans = (ans << 1) % mod;
	ans = (ans << 1) % mod;
	printf("%d\n",ans);
	return 0;
}

```