#include<bits/stdc++.h>
#define rep(i, l, r) for(int i = (l), i##end = (r);i <= i##end;++i)
using std::cin; using std::cout;
const int maxn = 3030;
typedef long long ll;
const int mod = 1e9 + 7;
inline void reduce(int & x) {
	x += x >> 31 & mod;
}
int n, m;
int f[maxn][maxn][2];
int main() {
	std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n >> m;
	rep(i, 0, n) f[0][i][!i] = 1;
	rep(i, 1, m) {
		rep(j, 0, n) {
			rep(k, 0, 1) {
				if(j != 0) {
					reduce(f[i][j][k || j == 1] += f[i - 1][j][k] - mod);
					reduce(f[i][j - 1][k || j == 1] += f[i - 1][j][k] - mod);
				}
				if(j != n) {
					reduce(f[i][j][k] += f[i - 1][j][k] - mod);
					reduce(f[i][j + 1][k] += f[i - 1][j][k] - mod);
				}
			}
		}
	}
	int ans = 0;
	rep(i, 0, n) reduce(ans += f[m][i][1] - mod);
	cout << ans << '\n';
}

### 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>
#define rep(i, l, r) for(int i = (l), i##end = (r);i <= i##end;++i)
using std::cin; using std::cout;
const int maxn = 3030;
typedef long long ll;
const int mod = 1e9 + 7;
inline void reduce(int & x) {
	x += x >> 31 & mod;
}
int n, m;
int f[maxn][maxn][2];
int main() {
	std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n >> m;
	rep(i, 0, n) f[0][i][!i] = 1;
	rep(i, 1, m) {
		rep(j, 0, n) {
			rep(k, 0, 1) {
				if(j != 0) {
					reduce(f[i][j][k || j == 1] += f[i - 1][j][k] - mod);
					reduce(f[i][j - 1][k || j == 1] += f[i - 1][j][k] - mod);
				}
				if(j != n) {
					reduce(f[i][j][k] += f[i - 1][j][k] - mod);
					reduce(f[i][j + 1][k] += f[i - 1][j][k] - mod);
				}
			}
		}
	}
	int ans = 0;
	rep(i, 0, n) reduce(ans += f[m][i][1] - mod);
	cout << ans << '\n';
}

```