#include <bits/stdc++.h>

using namespace std;

using ll = long long;

const ll MOD = 1e9 + 7;

ll dp[3005][3005], ans[3005];

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n, m;
	cin >> n >> m;
	ll ans = 0;
	for (int lim = n; lim >= n - 1; lim--) {
		for (int i = 0; i <= lim; i++) dp[0][i] = 1;
		for (int i = 1; i <= m; i++) {
			for (int j = 0; j <= lim; j++) {
				dp[i][j] = 0;
			}
		}
		for (int i = 0; i < m; i++) {
			for (int j = 0; j <= lim; j++) {
				if (j > 0) {
					(dp[i + 1][j] += dp[i][j]) %= MOD;
					(dp[i + 1][j - 1] += dp[i][j]) %= MOD;
				}
				if (j < lim) {
					(dp[i + 1][j] += dp[i][j]) %= MOD;
					(dp[i + 1][j + 1] += dp[i][j]) %= MOD;
				}
			}
		}	

		for (int i = 0; i <= lim; i++) {
			if (lim == n) (ans += dp[m][i]) %= MOD;
			else (ans += (MOD - dp[m][i]) % MOD) %= MOD;
		}	
	}
	cout << ans << endl;
	return 0;
}
### Prompt

Please formulate 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;

using ll = long long;

const ll MOD = 1e9 + 7;

ll dp[3005][3005], ans[3005];

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n, m;
	cin >> n >> m;
	ll ans = 0;
	for (int lim = n; lim >= n - 1; lim--) {
		for (int i = 0; i <= lim; i++) dp[0][i] = 1;
		for (int i = 1; i <= m; i++) {
			for (int j = 0; j <= lim; j++) {
				dp[i][j] = 0;
			}
		}
		for (int i = 0; i < m; i++) {
			for (int j = 0; j <= lim; j++) {
				if (j > 0) {
					(dp[i + 1][j] += dp[i][j]) %= MOD;
					(dp[i + 1][j - 1] += dp[i][j]) %= MOD;
				}
				if (j < lim) {
					(dp[i + 1][j] += dp[i][j]) %= MOD;
					(dp[i + 1][j + 1] += dp[i][j]) %= MOD;
				}
			}
		}	

		for (int i = 0; i <= lim; i++) {
			if (lim == n) (ans += dp[m][i]) %= MOD;
			else (ans += (MOD - dp[m][i]) % MOD) %= MOD;
		}	
	}
	cout << ans << endl;
	return 0;
}
```