#include <bits/stdc++.h>
using namespace std;
const int maxn = 3010;
const int mod = 1e9+7;
bool vis[maxn][maxn][2];
int memo[maxn][maxn][2];
int n, m;

int add(int a, int b) {return (a+b)%mod;}
int mul(long long a, long long b) {return a*b%mod;}

int dp(int pos, int r, int b, bool k) {
	if (pos == m) return k;
	if (vis[pos][r][k]) return memo[pos][r][k];
	int &ans = memo[pos][r][k] = 0;
	if (r > 0) {
		ans = add(ans, dp(pos+1, r-1, b+1, k|(r==1)));
		ans = add(ans, dp(pos+1, r, b, k|(r==1)));
	}
	if (b > 0) {
		ans = add(ans, dp(pos+1, r+1, b-1, k|(r==0)));
		ans = add(ans, dp(pos+1, r, b, k|(r==0)));
	}
	vis[pos][r][k] = 1;
	return ans;
}

int main() {
	cin >> n >> m;
	int ans = 0;
	for (int i = 0; i <= n; ++i) {
		ans = add(ans, dp(0, i, n-i, 0));
	}
	cout << ans << endl;
	return 0;
}
### Prompt

Please create a solution in CPP 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;
bool vis[maxn][maxn][2];
int memo[maxn][maxn][2];
int n, m;

int add(int a, int b) {return (a+b)%mod;}
int mul(long long a, long long b) {return a*b%mod;}

int dp(int pos, int r, int b, bool k) {
	if (pos == m) return k;
	if (vis[pos][r][k]) return memo[pos][r][k];
	int &ans = memo[pos][r][k] = 0;
	if (r > 0) {
		ans = add(ans, dp(pos+1, r-1, b+1, k|(r==1)));
		ans = add(ans, dp(pos+1, r, b, k|(r==1)));
	}
	if (b > 0) {
		ans = add(ans, dp(pos+1, r+1, b-1, k|(r==0)));
		ans = add(ans, dp(pos+1, r, b, k|(r==0)));
	}
	vis[pos][r][k] = 1;
	return ans;
}

int main() {
	cin >> n >> m;
	int ans = 0;
	for (int i = 0; i <= n; ++i) {
		ans = add(ans, dp(0, i, n-i, 0));
	}
	cout << ans << endl;
	return 0;
}
```