#include <bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef pair<int, int> pi;
const int mod = 1e9 + 7;

int dp[3005][3005][2], n, m;

int f(int x, int g, int p){
	if(x == m) return p;
	if(~dp[x][g][p]) return dp[x][g][p];
	lint ret = 0;
	if(g < n) ret += f(x+1, g+1, p);
	if(g > 0) ret += f(x+1, g, p | (g == 1));
	if(g < n) ret += f(x+1, g, p);
	if(g > 0) ret += f(x+1, g-1, p | (g == 1));
	return dp[x][g][p] = ret % mod;
}

int main(){
	cin >> n >> m;
	memset(dp, -1, sizeof(dp));
	lint ret = 0;
	for(int i=0; i<=n; i++) ret += f(0, i, i == 0);
	cout << ret % mod;
}

### Prompt

Your challenge is to write 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;
typedef long long lint;
typedef pair<int, int> pi;
const int mod = 1e9 + 7;

int dp[3005][3005][2], n, m;

int f(int x, int g, int p){
	if(x == m) return p;
	if(~dp[x][g][p]) return dp[x][g][p];
	lint ret = 0;
	if(g < n) ret += f(x+1, g+1, p);
	if(g > 0) ret += f(x+1, g, p | (g == 1));
	if(g < n) ret += f(x+1, g, p);
	if(g > 0) ret += f(x+1, g-1, p | (g == 1));
	return dp[x][g][p] = ret % mod;
}

int main(){
	cin >> n >> m;
	memset(dp, -1, sizeof(dp));
	lint ret = 0;
	for(int i=0; i<=n; i++) ret += f(0, i, i == 0);
	cout << ret % mod;
}

```