#include <bits/stdc++.h>
const int N=3050,mo=1e9+7;
using namespace std;
inline int gi(){
  char ch=getchar();int x=0,q=0;
  while(ch<'0' || ch>'9') ch=='-'?q=1:0,ch=getchar();
  while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
  return q?(-x):x;
}
int f[N][N][2];
void dp(int &x,int y){
  x+=y; if (x>=mo) x-=mo;
  return;
}
int main(){
  int n=gi(),m=gi();
  for (int i=0; i<=m; ++i)
    f[0][i][i==0]=1;
  for (int i=0; i<m; ++i)
    for (int j=0; j<=n; ++j)
      for (int k=0; k<=1; ++k)
      if (f[i][j][k]){
	if (j) dp(f[i+1][j-1][k|(j==1)],f[i][j][k]);
	if (j) dp(f[i+1][j][k|(j==1)],f[i][j][k]);
	if (j<n) dp(f[i+1][j+1][k],f[i][j][k]);
	if (j<n) dp(f[i+1][j][k],f[i][j][k]);
      }
  int ans=0;
  for (int i=0; i<=n; ++i)
    dp(ans,f[m][i][1]);
  cout<<ans;
  return 0;
}

### Prompt

Please formulate 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>
const int N=3050,mo=1e9+7;
using namespace std;
inline int gi(){
  char ch=getchar();int x=0,q=0;
  while(ch<'0' || ch>'9') ch=='-'?q=1:0,ch=getchar();
  while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
  return q?(-x):x;
}
int f[N][N][2];
void dp(int &x,int y){
  x+=y; if (x>=mo) x-=mo;
  return;
}
int main(){
  int n=gi(),m=gi();
  for (int i=0; i<=m; ++i)
    f[0][i][i==0]=1;
  for (int i=0; i<m; ++i)
    for (int j=0; j<=n; ++j)
      for (int k=0; k<=1; ++k)
      if (f[i][j][k]){
	if (j) dp(f[i+1][j-1][k|(j==1)],f[i][j][k]);
	if (j) dp(f[i+1][j][k|(j==1)],f[i][j][k]);
	if (j<n) dp(f[i+1][j+1][k],f[i][j][k]);
	if (j<n) dp(f[i+1][j][k],f[i][j][k]);
      }
  int ans=0;
  for (int i=0; i<=n; ++i)
    dp(ans,f[m][i][1]);
  cout<<ans;
  return 0;
}

```