#include <bits/stdc++.h>
using namespace std;
using ll=long long;
const ll mn=3007;
const ll mod=1e9+7;
ll dp[mn][mn][2];

void modAdd(ll &a,ll b){
	a+=b;
	if(a>=mod)a%=mod;
}

int main(){
	ll N,M;cin>>N>>M;
	for(ll a=0;a<=N;++a)dp[0][a][a==0]=1;
	for(ll a=0;a<M;++a){
		for(ll b=0;b<=N;++b){
			for(ll c=0;c<2;++c){
				if(dp[a][b][c]<1)continue;
				ll d=dp[a][b][c];
				if(b>0)modAdd(dp[a+1][b][c|b==1],d);
				if(b>0)modAdd(dp[a+1][b-1][c|b==1],d);
				if(b<N)modAdd(dp[a+1][b][c],d);
				if(b<N)modAdd(dp[a+1][b+1][c],d);
			}
		}
	}
	ll ret;
	for(ll a=0;a<=N;++a)modAdd(ret,dp[M][a][1]);
	cout<<ret<<endl;
}

### Prompt

In Cpp, your task is to solve 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;
using ll=long long;
const ll mn=3007;
const ll mod=1e9+7;
ll dp[mn][mn][2];

void modAdd(ll &a,ll b){
	a+=b;
	if(a>=mod)a%=mod;
}

int main(){
	ll N,M;cin>>N>>M;
	for(ll a=0;a<=N;++a)dp[0][a][a==0]=1;
	for(ll a=0;a<M;++a){
		for(ll b=0;b<=N;++b){
			for(ll c=0;c<2;++c){
				if(dp[a][b][c]<1)continue;
				ll d=dp[a][b][c];
				if(b>0)modAdd(dp[a+1][b][c|b==1],d);
				if(b>0)modAdd(dp[a+1][b-1][c|b==1],d);
				if(b<N)modAdd(dp[a+1][b][c],d);
				if(b<N)modAdd(dp[a+1][b+1][c],d);
			}
		}
	}
	ll ret;
	for(ll a=0;a<=N;++a)modAdd(ret,dp[M][a][1]);
	cout<<ret<<endl;
}

```