#include <iostream>
#include <vector>

using namespace std;

const int64_t MOD = 1e9 + 7;

int N, M;

vector<vector<vector<int64_t>>> DP(2, vector<vector<int64_t>>(3001, vector<int64_t>(3001, -1)));

int64_t f(bool bottle, int r, int steps) {
	if (steps == 0)
		return bottle;
	int64_t &ans = DP[bottle][r][steps];
	if (ans != -1)
		return ans;
	ans = 0;
	if (r > 0) {
		ans += f(bottle || r == 1, r - 1, steps - 1);
		ans += f(bottle || r == 1, r, steps - 1);
	}
	if (r < N) {
		ans += f(bottle || r == 0, r, steps - 1);
		ans += f(bottle || r == 0, r+1, steps - 1);
	}
	ans %= MOD;
	return ans;
}

int main() {
	cin >> N >> M;
	int64_t ways = 0;
	for (int r = 0; r <= N; ++r) {
		ways += f(0, r, M);
		ways %= MOD;
	}
	cout << ways << endl;
	cin >> ws;
}
### Prompt

Develop a solution in CPP to the problem described below:

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 <iostream>
#include <vector>

using namespace std;

const int64_t MOD = 1e9 + 7;

int N, M;

vector<vector<vector<int64_t>>> DP(2, vector<vector<int64_t>>(3001, vector<int64_t>(3001, -1)));

int64_t f(bool bottle, int r, int steps) {
	if (steps == 0)
		return bottle;
	int64_t &ans = DP[bottle][r][steps];
	if (ans != -1)
		return ans;
	ans = 0;
	if (r > 0) {
		ans += f(bottle || r == 1, r - 1, steps - 1);
		ans += f(bottle || r == 1, r, steps - 1);
	}
	if (r < N) {
		ans += f(bottle || r == 0, r, steps - 1);
		ans += f(bottle || r == 0, r+1, steps - 1);
	}
	ans %= MOD;
	return ans;
}

int main() {
	cin >> N >> M;
	int64_t ways = 0;
	for (int r = 0; r <= N; ++r) {
		ways += f(0, r, M);
		ways %= MOD;
	}
	cout << ways << endl;
	cin >> ws;
}
```