#include<bits/stdc++.h>
#define all(x) x.begin(), x.end()
using namespace std;
using ll = long long;
const int maxn = 3030, mod = 1e9 + 7;
int n, m, dp[maxn][2*maxn][2];
inline void add(int &a, int b) {
	a = a+b >= mod ? a+b-mod : a+b;
}
int main() {
	cin.tie(0)->sync_with_stdio(0);
	cin >> n >> m;
	for(int i = 0; i <= n; i++) dp[0][i][i==0] = 1;
	for(int i = 0; i < m; i++) {
		for(int j = 0; j <= n; j++) {
			int k = n-j;
			for(int f = 0; f < 2; f++) {
				if(j) {
					add(dp[i+1][j-1][f|(j==1)], dp[i][j][f]);
					add(dp[i+1][j][f|(j==1)], dp[i][j][f]);
				}
				if(k) {
					add(dp[i+1][j][f], dp[i][j][f]);
					add(dp[i+1][j+1][f], dp[i][j][f]);
				}
			}
		}
	}
	int ans = 0;
	for(int i = 0; i <= n; i++) add(ans, dp[m][i][1]);
	cout << ans;
	return 0;
}

### Prompt

Your task is to create 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>
#define all(x) x.begin(), x.end()
using namespace std;
using ll = long long;
const int maxn = 3030, mod = 1e9 + 7;
int n, m, dp[maxn][2*maxn][2];
inline void add(int &a, int b) {
	a = a+b >= mod ? a+b-mod : a+b;
}
int main() {
	cin.tie(0)->sync_with_stdio(0);
	cin >> n >> m;
	for(int i = 0; i <= n; i++) dp[0][i][i==0] = 1;
	for(int i = 0; i < m; i++) {
		for(int j = 0; j <= n; j++) {
			int k = n-j;
			for(int f = 0; f < 2; f++) {
				if(j) {
					add(dp[i+1][j-1][f|(j==1)], dp[i][j][f]);
					add(dp[i+1][j][f|(j==1)], dp[i][j][f]);
				}
				if(k) {
					add(dp[i+1][j][f], dp[i][j][f]);
					add(dp[i+1][j+1][f], dp[i][j][f]);
				}
			}
		}
	}
	int ans = 0;
	for(int i = 0; i <= n; i++) add(ans, dp[m][i][1]);
	cout << ans;
	return 0;
}

```