#include<bits/stdc++.h>
const int P = 1e9 + 7;
int dp[3005][3005][2];

int main() {
  if(fopen("input.txt","r")) {
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
  }

  int N, M; scanf("%d%d", &N, &M);
  for(int i = 1;i <= N; ++i)
    dp[0][i][0] = 1;
  dp[0][0][1] = 1;
  for(int i = 0;i < M; ++i) {
    for(int j = 0;j <= N; ++j) {
      for(int k = 0;k < 2; ++k) {
	if(!dp[i][j][k]) continue;
	if(j) {
	  (dp[i+1][j-1][k|(j==1)] += dp[i][j][k]) %= P;
	  (dp[i+1][j][k|(j==1)] += dp[i][j][k]) %= P;
	}
	if(N-j) {
	  (dp[i+1][j+1][k] += dp[i][j][k]) %= P;
	  (dp[i+1][j][k] += dp[i][j][k]) %= P;
	}
      }
    }
  }

  int ans = 0;
  for(int i = 0;i <= N; ++i)
    (ans += dp[M][i][1]) %= P;
  printf("%d", ans);
  
  return 0;
}

### Prompt

Your task is to create 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 P = 1e9 + 7;
int dp[3005][3005][2];

int main() {
  if(fopen("input.txt","r")) {
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
  }

  int N, M; scanf("%d%d", &N, &M);
  for(int i = 1;i <= N; ++i)
    dp[0][i][0] = 1;
  dp[0][0][1] = 1;
  for(int i = 0;i < M; ++i) {
    for(int j = 0;j <= N; ++j) {
      for(int k = 0;k < 2; ++k) {
	if(!dp[i][j][k]) continue;
	if(j) {
	  (dp[i+1][j-1][k|(j==1)] += dp[i][j][k]) %= P;
	  (dp[i+1][j][k|(j==1)] += dp[i][j][k]) %= P;
	}
	if(N-j) {
	  (dp[i+1][j+1][k] += dp[i][j][k]) %= P;
	  (dp[i+1][j][k] += dp[i][j][k]) %= P;
	}
      }
    }
  }

  int ans = 0;
  for(int i = 0;i <= N; ++i)
    (ans += dp[M][i][1]) %= P;
  printf("%d", ans);
  
  return 0;
}

```