#include<bits/stdc++.h>
#define N 3002
using namespace std;
typedef long long ll;
const int mod=1e9+7;
ll dp[N][N][2];
int n,m;
inline void MOD(ll &x){x=x>=mod?x-mod:x;}
inline ll rd(){
  ll x=0;char c=getchar();bool f=0;
  while(!isdigit(c)){if(c=='-')f=1;c=getchar();}
  while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();}
  return f?-x:x;
}
int main(){
  n=rd();m=rd();
  for(int i=1;i<=n;++i)dp[0][i][0]=1;
  dp[0][0][1]=1;
  for(int i=1;i<=m;++i){
  	for(int j=0;j<=n;++j)
  		for(int k=0;k<2;++k)if(dp[i-1][j][k]){
  		  if(j){
  		  	MOD(dp[i][j-1][k|(j==1)]+=dp[i-1][j][k]);
  		  	MOD(dp[i][j][k|(j==1)]+=dp[i-1][j][k]);
  		  }
          if(j<n){
          	MOD(dp[i][j+1][k]+=dp[i-1][j][k]);
            MOD(dp[i][j][k]+=dp[i-1][j][k]);
          }
  	    }
  }
  ll ans=0;
  for(int i=0;i<=n;++i)MOD(ans+=dp[m][i][1]);
  cout<<ans;
  return 0;
}

### 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 N 3002
using namespace std;
typedef long long ll;
const int mod=1e9+7;
ll dp[N][N][2];
int n,m;
inline void MOD(ll &x){x=x>=mod?x-mod:x;}
inline ll rd(){
  ll x=0;char c=getchar();bool f=0;
  while(!isdigit(c)){if(c=='-')f=1;c=getchar();}
  while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();}
  return f?-x:x;
}
int main(){
  n=rd();m=rd();
  for(int i=1;i<=n;++i)dp[0][i][0]=1;
  dp[0][0][1]=1;
  for(int i=1;i<=m;++i){
  	for(int j=0;j<=n;++j)
  		for(int k=0;k<2;++k)if(dp[i-1][j][k]){
  		  if(j){
  		  	MOD(dp[i][j-1][k|(j==1)]+=dp[i-1][j][k]);
  		  	MOD(dp[i][j][k|(j==1)]+=dp[i-1][j][k]);
  		  }
          if(j<n){
          	MOD(dp[i][j+1][k]+=dp[i-1][j][k]);
            MOD(dp[i][j][k]+=dp[i-1][j][k]);
          }
  	    }
  }
  ll ans=0;
  for(int i=0;i<=n;++i)MOD(ans+=dp[m][i][1]);
  cout<<ans;
  return 0;
}

```