#include <bits/stdc++.h>
using namespace std;

const int64_t MOD = 1e9+7;
void add(int64_t& a, int64_t b){
    a = (a+b) % MOD;
}
int64_t dp[3001][3001][2];

int main(){
    int i, j, k;
    int N, M;
    cin >> N >> M;

    for(j=0; j<=N; j++) dp[0][j][(j==0)] = 1;
    for(i=0; i<M; i++){
        for(j=0; j<=N; j++){
            for(k=0; k<2; k++){
                if(j>0){
                    add(dp[i+1][j][k|(j==1)], dp[i][j][k]);
                    add(dp[i+1][j-1][k|(j==1)], dp[i][j][k]);
                }
                if(j<N){
                    add(dp[i+1][j+1][k], dp[i][j][k]);
                    add(dp[i+1][j][k], dp[i][j][k]);
                }
            }
        }
    }

    int64_t ans = 0;
    for(j=0; j<=N; j++) add(ans, dp[M][j][1]);
    cout << ans << endl;
    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>
using namespace std;

const int64_t MOD = 1e9+7;
void add(int64_t& a, int64_t b){
    a = (a+b) % MOD;
}
int64_t dp[3001][3001][2];

int main(){
    int i, j, k;
    int N, M;
    cin >> N >> M;

    for(j=0; j<=N; j++) dp[0][j][(j==0)] = 1;
    for(i=0; i<M; i++){
        for(j=0; j<=N; j++){
            for(k=0; k<2; k++){
                if(j>0){
                    add(dp[i+1][j][k|(j==1)], dp[i][j][k]);
                    add(dp[i+1][j-1][k|(j==1)], dp[i][j][k]);
                }
                if(j<N){
                    add(dp[i+1][j+1][k], dp[i][j][k]);
                    add(dp[i+1][j][k], dp[i][j][k]);
                }
            }
        }
    }

    int64_t ans = 0;
    for(j=0; j<=N; j++) add(ans, dp[M][j][1]);
    cout << ans << endl;
    return 0;
}
```