#include <bits/stdc++.h>

using namespace std;

typedef long long int64;
const int mod = 1e9 + 7;

int64 N, M;
int64 dp[3001][3001][2];

int64 rec(int step, int idx, bool touch)
{
  if(step == M) return (touch);
  touch |= idx == 0 || idx == N;
  if(~dp[step][idx][touch]) return (dp[step][idx][touch]);
  int ret = 0;
  if(idx - 1 >= 0) (ret += rec(step + 1, idx, touch)) %= mod;
  if(idx + 1 <= N) (ret += rec(step + 1, idx, touch)) %= mod;
  if(idx - 1 >= 0) (ret += rec(step + 1, idx - 1, touch)) %= mod;
  if(idx + 1 <= N) (ret += rec(step + 1, idx + 1, touch)) %= mod;
  return (dp[step][idx][touch] = ret);
}


int main()
{
  memset(dp, -1, sizeof(dp));
  cin >> N >> M;
  int ret = 0;
  for(int i = 0; i <= N; i++) (ret += rec(0, i, false)) %= mod;
  cout << ret << endl;
}

### 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>

using namespace std;

typedef long long int64;
const int mod = 1e9 + 7;

int64 N, M;
int64 dp[3001][3001][2];

int64 rec(int step, int idx, bool touch)
{
  if(step == M) return (touch);
  touch |= idx == 0 || idx == N;
  if(~dp[step][idx][touch]) return (dp[step][idx][touch]);
  int ret = 0;
  if(idx - 1 >= 0) (ret += rec(step + 1, idx, touch)) %= mod;
  if(idx + 1 <= N) (ret += rec(step + 1, idx, touch)) %= mod;
  if(idx - 1 >= 0) (ret += rec(step + 1, idx - 1, touch)) %= mod;
  if(idx + 1 <= N) (ret += rec(step + 1, idx + 1, touch)) %= mod;
  return (dp[step][idx][touch] = ret);
}


int main()
{
  memset(dp, -1, sizeof(dp));
  cin >> N >> M;
  int ret = 0;
  for(int i = 0; i <= N; i++) (ret += rec(0, i, false)) %= mod;
  cout << ret << endl;
}

```