#include <bits/stdc++.h>

const int MAXN = 3010;
const int mod = 1000000007;
void reduce(int & x) { x += x >> 31 & mod; }
int f[2][MAXN][2], n, m;
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	std::cin >> n >> m;
	int ans = 0, now = 1, lst = 0;
	for (int i = 0; i <= n; ++i) f[now][i][i == 0] = 1;
	for (int i = 0; i < m * 2; ++i) {
		std::swap(now, lst);
		memset(f[now], 0, sizeof f[now]);
		int ax = i % 2 == 0, up = n + (!ax);
		for (int j = 0; j <= up; ++j) 
		for (int k = 0; k != 2; ++k) if (int t = f[lst][j][k]) {
			if (j)
			reduce(f[now][j - 1 + ax][k | (j == 1)] += t - mod);
			if (j < up)
			reduce(f[now][j + ax][k] += t - mod);
		}
	}
	for (int i = 0; i <= n; ++i) reduce(ans += f[now][i][1] - mod);
	std::cout << ans << std::endl;
	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>

const int MAXN = 3010;
const int mod = 1000000007;
void reduce(int & x) { x += x >> 31 & mod; }
int f[2][MAXN][2], n, m;
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	std::cin >> n >> m;
	int ans = 0, now = 1, lst = 0;
	for (int i = 0; i <= n; ++i) f[now][i][i == 0] = 1;
	for (int i = 0; i < m * 2; ++i) {
		std::swap(now, lst);
		memset(f[now], 0, sizeof f[now]);
		int ax = i % 2 == 0, up = n + (!ax);
		for (int j = 0; j <= up; ++j) 
		for (int k = 0; k != 2; ++k) if (int t = f[lst][j][k]) {
			if (j)
			reduce(f[now][j - 1 + ax][k | (j == 1)] += t - mod);
			if (j < up)
			reduce(f[now][j + ax][k] += t - mod);
		}
	}
	for (int i = 0; i <= n; ++i) reduce(ans += f[now][i][1] - mod);
	std::cout << ans << std::endl;
	return 0;
}

```