#include <bits/stdc++.h>

using namespace std;

const int maxn = 3100;
const int base = 1e9+7;

int n,m,f[maxn][maxn][3];

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin>>n>>m;
    f[0][0][1]=1;
    for (int j=1;j<=n;j++)
        f[0][j][0]=1;
    for (int i=0;i<m;i++)
        for (int j=0;j<=n;j++)
            for (int tt =0;tt<2;tt++) {
                if (j) {
                    f[i+1][j][tt | (j==1)] = (f[i+1][j][tt | (j==1)] + f[i][j][tt]) % base;
                    f[i+1][j-1][tt | (j==1)] = (f[i+1][j-1][tt | (j==1)] + f[i][j][tt]) % base;
                }
                if (j<n) {
                    f[i+1][j][tt | (j==0)] = (f[i+1][j][tt | (j==0)] + f[i][j][tt]) % base;
                    f[i+1][j+1][tt] = (f[i+1][j+1][tt] + f[i][j][tt]) % base;
                }
            }
    int res=0;
    for (int j=0;j<=n;j++) res = (res + f[m][j][1]) % base;
    cout<<res<<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;

const int maxn = 3100;
const int base = 1e9+7;

int n,m,f[maxn][maxn][3];

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin>>n>>m;
    f[0][0][1]=1;
    for (int j=1;j<=n;j++)
        f[0][j][0]=1;
    for (int i=0;i<m;i++)
        for (int j=0;j<=n;j++)
            for (int tt =0;tt<2;tt++) {
                if (j) {
                    f[i+1][j][tt | (j==1)] = (f[i+1][j][tt | (j==1)] + f[i][j][tt]) % base;
                    f[i+1][j-1][tt | (j==1)] = (f[i+1][j-1][tt | (j==1)] + f[i][j][tt]) % base;
                }
                if (j<n) {
                    f[i+1][j][tt | (j==0)] = (f[i+1][j][tt | (j==0)] + f[i][j][tt]) % base;
                    f[i+1][j+1][tt] = (f[i+1][j+1][tt] + f[i][j][tt]) % base;
                }
            }
    int res=0;
    for (int j=0;j<=n;j++) res = (res + f[m][j][1]) % base;
    cout<<res<<endl;
}

```