#include <iostream>
#include <vector>

using namespace std;

const int MOD = 1000000007;

int main(){
	int N, M;
	while(cin >> N >> M){
		vector<vector<int>> init(2, vector<int>(N+1, 0));
		auto dp = init;
		for(int i=0;i<=N;i++) dp[i==0?1:0][i] = 1;
		for(int turn=0;turn<M;turn++){
			auto next = init;
			for(int j=0;j<2;j++){
				for(int i=0;i<=N;i++){
					if(i > 0){
						next[i==1?1:j][i-1] = (next[i==1?1:j][i-1]+dp[j][i])%MOD;
						next[i==1?1:j][i  ] = (next[i==1?1:j][i  ]+dp[j][i])%MOD;
					}
					if(i < N){
						next[j][i  ] = (next[j][i  ]+dp[j][i])%MOD;
						next[j][i+1] = (next[j][i+1]+dp[j][i])%MOD;
					}
				}
			}
			dp = next;
		}
		int res = 0;
		for(int i=0;i<=N;i++) res = (res+dp[1][i])%MOD;
		cout << res << endl;
	}
}

### Prompt

Create a solution in CPP for 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 <iostream>
#include <vector>

using namespace std;

const int MOD = 1000000007;

int main(){
	int N, M;
	while(cin >> N >> M){
		vector<vector<int>> init(2, vector<int>(N+1, 0));
		auto dp = init;
		for(int i=0;i<=N;i++) dp[i==0?1:0][i] = 1;
		for(int turn=0;turn<M;turn++){
			auto next = init;
			for(int j=0;j<2;j++){
				for(int i=0;i<=N;i++){
					if(i > 0){
						next[i==1?1:j][i-1] = (next[i==1?1:j][i-1]+dp[j][i])%MOD;
						next[i==1?1:j][i  ] = (next[i==1?1:j][i  ]+dp[j][i])%MOD;
					}
					if(i < N){
						next[j][i  ] = (next[j][i  ]+dp[j][i])%MOD;
						next[j][i+1] = (next[j][i+1]+dp[j][i])%MOD;
					}
				}
			}
			dp = next;
		}
		int res = 0;
		for(int i=0;i<=N;i++) res = (res+dp[1][i])%MOD;
		cout << res << endl;
	}
}

```